Welcome dear students! Today we are going to learn about Statistics from Class 10 Maths.
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In Class Nine, you studied the classification of given data into ungrouped as well as grouped frequency distributions. You also learnt to represent the data pictorially in the form of various graphs such as bar graphs, histograms including those of varying widths and frequency polygons. In fact, you went a step further by studying certain numerical representatives of the ungrouped data, also called measures of central tendency, namely, mean, median and mode. In this chapter, we shall extend the study of these three measures, that is, mean, median and mode from ungrouped data to that of grouped data.
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Let us begin with Section 13.2, Mean of Grouped Data. The mean, or average, of observations, as we know, is the sum of the values of all the observations divided by the total number of observations. From Class Nine, recall that if x1, x2, dot dot dot, xn are observations with respective frequencies f1, f2, dot dot dot, fn, then this means observation x1 occurs f1 times, x2 occurs f2 times, and so on. Now, the sum of the values of all the observations equals f1x1 plus f2x2 plus dot dot dot plus fnxn, and the number of observations equals f1 plus f2 plus dot dot dot plus fn. So, the mean x bar of the data is given by x bar equals f1x1 plus f2x2 plus dot dot dot plus fnxn divided by f1 plus f2 plus dot dot dot plus fn. Recall that we can write this in short form by using the Greek letter capital sigma which means summation. That is, x bar equals sigma from i equals one to n of fi xi divided by sigma from i equals one to n of fi, which, more briefly, is written as x bar equals sigma fi xi divided by sigma fi, if it is understood that i varies from one to n.
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Let us apply this formula to find the mean in the following example. Example 1: The marks obtained by 30 students of Class Ten of a certain school in a Mathematics paper consisting of 100 marks are presented in table below. Find the mean of the marks obtained by the students. The data is as follows: Marks obtained ten, twenty, thirty-six, forty, fifty, fifty-six, sixty, seventy, seventy-two, eighty, eighty-eight, ninety-two, ninety-five. The corresponding number of students are one, one, three, four, three, two, four, four, one, one, two, three, one. Solution: Recall that to find the mean marks, we require the product of each xi with the corresponding frequency fi. So, let us put them in a column. For marks ten, frequency is one, product is ten. For twenty, frequency one, product twenty. For thirty-six, frequency three, product one hundred eight. For forty, frequency four, product one hundred sixty. For fifty, frequency three, product one hundred fifty. For fifty-six, frequency two, product one hundred twelve. For sixty, frequency four, product two hundred forty. For seventy, frequency four, product two hundred eighty. For seventy-two, frequency one, product seventy-two. For eighty, frequency one, product eighty. For eighty-eight, frequency two, product one hundred seventy-six. For ninety-two, frequency three, product two hundred seventy-six. For ninety-five, frequency one, product ninety-five. The total frequency sigma fi is thirty. The total product sigma fi xi is one thousand seven hundred seventy-nine.
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Now, x bar equals sigma fi xi divided by sigma fi equals one thousand seven hundred seventy-nine divided by thirty equals fifty-nine point three. Therefore, the mean marks obtained is fifty-nine point three. In most of our real life situations, data is usually so large that to make a meaningful study it needs to be condensed as grouped data. So, we need to convert given ungrouped data into grouped data and devise some method to find its mean. Let us convert the ungrouped data of Example 1 into grouped data by forming class-intervals of width, say fifteen. Remember that, while allocating frequencies to each class-interval, students falling in any upper class-limit would be considered in the next class, for example, four students who have obtained forty marks would be considered in the class-interval forty to fifty-five and not in twenty-five to forty. With this convention in our mind, let us form a grouped frequency distribution table. The class intervals are ten to twenty-five, twenty-five to forty, forty to fifty-five, fifty-five to seventy, seventy to eighty-five, and eighty-five to one hundred. The number of students are two, three, seven, six, six, six.
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Now, for each class-interval, we require a point which would serve as the representative of the whole class. It is assumed that the frequency of each class-interval is centred around its mid-point. So the mid-point, or class mark, of each class can be chosen to represent the observations falling in the class. Recall that we find the mid-point of a class, or its class mark, by finding the average of its upper and lower limits. That is, Class mark equals Upper class limit plus Lower class limit divided by two. With reference to the table, for the class ten to twenty-five, the class mark is ten plus twenty-five divided by two, that is, seventeen point five. Similarly, we can find the class marks of the remaining class intervals. We put them in a table. These class marks serve as our xi values. For class ten to twenty-five, frequency is two, class mark seventeen point five, product thirty-five point zero. For twenty-five to forty, frequency three, class mark thirty-two point five, product ninety-seven point five. For forty to fifty-five, frequency seven, class mark forty-seven point five, product three hundred thirty-two point five. For fifty-five to seventy, frequency six, class mark sixty-two point five, product three hundred seventy-five point zero. For seventy to eighty-five, frequency six, class mark seventy-seven point five, product four hundred sixty-five point zero. For eighty-five to one hundred, frequency six, class mark ninety-two point five, product five hundred fifty-five point zero. Total sigma fi is thirty. Total sigma fi xi is one thousand eight hundred sixty point zero.
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The sum of the values in the last column gives us sigma fi xi. So, the mean x bar of the given data is given by x bar equals one thousand eight hundred sixty point zero divided by thirty equals sixty-two. This new method of finding the mean is known as the Direct Method. We observe that the previous table and this table are using the same data and employing the same formula for the calculation of the mean but the results obtained are different. Can you think why this is so, and which one is more accurate? The difference in the two values is because of the mid-point assumption in this table, fifty-nine point three being the exact mean, while sixty-two an approximate mean. Sometimes when the numerical values of xi and fi are large, finding the product of xi and fi becomes tedious and time consuming. So, for such situations, let us think of a method of reducing these calculations. We can do nothing with the fi values, but we can change each xi to a smaller number so that our calculations become easy. How do we do this? What about subtracting a fixed number from each of these xi values? Let us try this method. The first step is to choose one among the xi values as the assumed mean, and denote it by a. Also, to further reduce our calculation work, we may take a to be that xi which lies in the centre of x1, x2, dot dot dot, xn. So, we can choose a equals forty-seven point five or a equals sixty-two point five. Let us choose a equals forty-seven point five.
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The next step is to find the difference di between a and each of the xi values, that is, the deviation of a from each of the xi values. That is, di equals xi minus a equals xi minus forty-seven point five. The third step is to find the product of di with the corresponding fi, and take the sum of all the fi di values. For class ten to twenty-five, frequency two, class mark seventeen point five, di is minus thirty, fidi is minus sixty. For twenty-five to forty, frequency three, class mark thirty-two point five, di is minus fifteen, fidi is minus forty-five. For forty to fifty-five, frequency seven, class mark forty-seven point five, di is zero, fidi is zero. For fifty-five to seventy, frequency six, class mark sixty-two point five, di is fifteen, fidi is ninety. For seventy to eighty-five, frequency six, class mark seventy-seven point five, di is thirty, fidi is one hundred eighty. For eighty-five to one hundred, frequency six, class mark ninety-two point five, di is forty-five, fidi is two hundred seventy. Total sigma fi is thirty. Total sigma fidi is four hundred thirty-five. So, from this table, the mean of the deviations, d bar equals sigma fidi divided by sigma fi. Now, let us find the relation between d bar and x bar. Since in obtaining di, we subtracted a from each xi, so, in order to get the mean x bar, we need to add a to d bar. This can be explained mathematically as: Mean of deviations, d bar equals sigma fi di divided by sigma fi. So, d bar equals sigma fi (xi minus a) divided by sigma fi equals sigma fi xi divided by sigma fi minus sigma fi a divided by sigma fi equals x bar minus a. So, x bar equals a plus d bar. That is, x bar equals a plus sigma fidi divided by sigma fi.
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Substituting the values of a, sigma fidi and sigma fi from the table, we get x bar equals four hundred thirty-five divided by thirty plus forty-seven point five equals fourteen point five plus forty-seven point five equals sixty-two. Therefore, the mean of the marks obtained by the students is sixty-two. The method discussed above is called the Assumed Mean Method. Activity 1: From the table find the mean by taking each of xi, that is, seventeen point five, thirty-two point five, and so on as a. What do you observe? You will find that the mean determined in each case is the same, that is, sixty-two. Why? So, we can say that the value of the mean obtained does not depend on the choice of a. Observe that in the deviation table, the values in the deviation column are all multiples of fifteen. So, if we divide the values in the entire deviation column by fifteen, we would get smaller numbers to multiply with fi. Here, fifteen is the class size of each class interval. So, let ui equals xi minus a divided by h, where a is the assumed mean and h is the class size. Now, we calculate ui in this way and continue as before, that is, find fi ui and then sigma fi ui. Taking h equals fifteen, let us form a table.
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For class ten to twenty-five, fi is two, xi is seventeen point five, di is minus thirty, ui is minus two, fiui is minus four. For twenty-five to forty, fi is three, xi is thirty-two point five, di is minus fifteen, ui is minus one, fiui is minus three. For forty to fifty-five, fi is seven, xi is forty-seven point five, di is zero, ui is zero, fiui is zero. For fifty-five to seventy, fi is six, xi is sixty-two point five, di is fifteen, ui is one, fiui is six. For seventy to eighty-five, fi is six, xi is seventy-seven point five, di is thirty, ui is two, fiui is twelve. For eighty-five to one hundred, fi is six, xi is ninety-two point five, di is forty-five, ui is three, fiui is eighteen. Total sigma fi is thirty. Total sigma fiui is twenty-nine. Let u bar equals sigma fiui divided by sigma fi. Here, again let us find the relation between u bar and x bar. We have, ui equals xi minus a divided by h. Therefore, u bar equals sigma fi (xi minus a divided by h) divided by sigma fi equals one over h times [sigma fi xi divided by sigma fi minus a]. So, h u bar equals x bar minus a. That is, x bar equals a plus h u bar. So, x bar equals a plus h times sigma fiui divided by sigma fi.
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Now, substituting the values of a, h, sigma fiui and sigma fi from the table, we get x bar equals forty-seven point five plus fifteen times twenty-nine divided by thirty equals forty-seven point five plus fourteen point five equals sixty-two. So, the mean marks obtained by a student is sixty-two. The method discussed above is called the Step-deviation method. We note that the step-deviation method will be convenient to apply if all the di values have a common factor. The mean obtained by all the three methods is the same. The assumed mean method and step-deviation method are just simplified forms of the direct method. The formula x bar equals a plus h u bar still holds if a and h are not as given above, but are any non-zero numbers such that ui equals xi minus a divided by h. Let us apply these methods in another example.
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Example 2: The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories of India. Find the mean percentage of female teachers by all the three methods discussed in this section. The data is: Percentage of female teachers fifteen to twenty-five, twenty-five to thirty-five, thirty-five to forty-five, forty-five to fifty-five, fifty-five to sixty-five, sixty-five to seventy-five, seventy-five to eighty-five. Number of states or union territories are six, eleven, seven, four, four, two, one. Solution: Let us find the class marks, xi, of each class, and put them in a column. For fifteen to twenty-five, frequency six, xi twenty. For twenty-five to thirty-five, frequency eleven, xi thirty. For thirty-five to forty-five, frequency seven, xi forty. For forty-five to fifty-five, frequency four, xi fifty. For fifty-five to sixty-five, frequency four, xi sixty. For sixty-five to seventy-five, frequency two, xi seventy. For seventy-five to eighty-five, frequency one, xi eighty. Here we take a equals fifty, h equals ten, then di equals xi minus fifty and ui equals xi minus fifty divided by ten. We now find di and ui and put them in a table. For fifteen to twenty-five, fi six, xi twenty, di minus thirty, ui minus three, fixi one hundred twenty, fidi minus one hundred eighty, fiui minus eighteen. For twenty-five to thirty-five, fi eleven, xi thirty, di minus twenty, ui minus two, fixi three hundred thirty, fidi minus two hundred twenty, fiui minus twenty-two. For thirty-five to forty-five, fi seven, xi forty, di minus ten, ui minus one, fixi two hundred eighty, fidi minus seventy, fiui minus seven. For forty-five to fifty-five, fi four, xi fifty, di zero, ui zero, fixi two hundred, fidi zero, fiui zero. For fifty-five to sixty-five, fi four, xi sixty, di ten, ui one, fixi two hundred forty, fidi forty, fiui four. For sixty-five to seventy-five, fi two, xi seventy, di twenty, ui two, fixi one hundred forty, fidi forty, fiui four. For seventy-five to eighty-five, fi one, xi eighty, di thirty, ui three, fixi eighty, fidi thirty, fiui three. Total sigma fi is thirty-five, sigma fixi is one thousand three hundred ninety, sigma fidi is minus three hundred sixty, sigma fiui is minus thirty-six.
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Using the direct method, x bar equals sigma fixi divided by sigma fi equals one thousand three hundred ninety divided by thirty-five equals thirty-nine point seven one. Using the assumed mean method, x bar equals a plus sigma fidi divided by sigma fi equals fifty plus minus three hundred sixty divided by thirty-five equals thirty-nine point seven one. Using the step-deviation method, x bar equals a plus h times sigma fiui divided by sigma fi equals fifty plus ten times minus thirty-six divided by thirty-five equals thirty-nine point seven one. Therefore, the mean percentage of female teachers in the primary schools of rural areas is thirty-nine point seven one. Remark: The result obtained by all the three methods is the same. So the choice of method to be used depends on the numerical values of xi and fi. If xi and fi are sufficiently small, then the direct method is an appropriate choice. If xi and fi are numerically large numbers, then we can go for the assumed mean method or step-deviation method. If the class sizes are unequal, and xi are large numerically, we can still apply the step-deviation method by taking h to be a suitable divisor of all the di values.
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Example 3: The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify? Number of wickets: twenty to sixty, sixty to one hundred, one hundred to one hundred fifty, one hundred fifty to two hundred fifty, two hundred fifty to three hundred fifty, three hundred fifty to four hundred fifty. Number of bowlers: seven, five, sixteen, twelve, two, three. Solution: Here, the class size varies, and the xi values are large. Let us still apply the step-deviation method with a equals two hundred and h equals twenty. Then, we obtain the data as follows. For twenty to sixty, fi seven, xi forty, di minus one hundred sixty, ui minus eight, fiui minus fifty-six. For sixty to one hundred, fi five, xi eighty, di minus one hundred twenty, ui minus six, fiui minus thirty. For one hundred to one hundred fifty, fi sixteen, xi one hundred twenty-five, di minus seventy-five, ui minus three point seven five, fiui minus sixty. For one hundred fifty to two hundred fifty, fi twelve, xi two hundred, di zero, ui zero, fiui zero. For two hundred fifty to three hundred fifty, fi two, xi three hundred, di one hundred, ui five, fiui ten. For three hundred fifty to four hundred fifty, fi three, xi four hundred, di two hundred, ui ten, fiui thirty. Total fi is forty-five. Total fiui is minus one hundred six. So, u bar equals minus one hundred six divided by forty-five. Therefore, x bar equals two hundred plus twenty times minus one hundred six divided by forty-five equals two hundred minus forty-seven point one one equals one hundred fifty-two point eight nine. This tells us that, on an average, the number of wickets taken by these forty-five bowlers in one-day cricket is one hundred fifty-two point eight nine.
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Now, let us see how well you can apply the concepts discussed in this section! Activity 2: Divide the students of your class into three groups and ask each group to do one of the following activities. One: Collect the marks obtained by all the students of your class in Mathematics in the latest examination conducted by your school. Form a grouped frequency distribution of the data obtained. Two: Collect the daily maximum temperatures recorded for a period of thirty days in your city. Present this data as a grouped frequency table. Three: Measure the heights of all the students of your class in centimetres and form a grouped frequency distribution table of this data. After all the groups have collected the data and formed grouped frequency distribution tables, the groups should find the mean in each case by the method which they find appropriate.
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Let us move to Exercise 13.1. Question 1: A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in twenty houses in a locality. Find the mean number of plants per house. Number of plants: zero to two, two to four, four to six, six to eight, eight to ten, ten to twelve, twelve to fourteen. Number of houses: one, two, one, five, six, two, three. Which method did you use for finding the mean, and why? Solution: We calculate class marks xi: one, three, five, seven, nine, eleven, thirteen. Frequencies fi: one, two, one, five, six, two, three. Products fi xi: one, six, five, thirty-five, fifty-four, twenty-two, thirty-nine. Sum of fi is twenty. Sum of fi xi is one hundred sixty-two. Mean x bar equals one hundred sixty-two divided by twenty equals eight point one. We used the direct method because the numerical values of xi and fi are small.
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Question 2: Consider the following distribution of daily wages of fifty workers of a factory. Daily wages in rupees: five hundred to five hundred twenty, five hundred twenty to five hundred forty, five hundred forty to five hundred sixty, five hundred sixty to five hundred eighty, five hundred eighty to six hundred. Number of workers: twelve, fourteen, eight, six, ten. Find the mean daily wages of the workers of the factory by using an appropriate method. Solution: Class marks xi: five hundred ten, five hundred thirty, five hundred fifty, five hundred seventy, five hundred ninety. Let us use the assumed mean method with a equals five hundred fifty. Deviations di: minus forty, minus twenty, zero, twenty, forty. fi di: minus four hundred eighty, minus two hundred eighty, zero, one hundred twenty, four hundred. Sum of fi is fifty. Sum of fi di is minus two hundred forty. Mean x bar equals five hundred fifty plus minus two hundred forty divided by fifty equals five hundred fifty minus four point eight equals five hundred forty-five point two. The mean daily wage is five hundred forty-five point two rupees.
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Question 3: The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f. Daily pocket allowance in rupees: eleven to thirteen, thirteen to fifteen, fifteen to seventeen, seventeen to nineteen, nineteen to twenty-one, twenty-one to twenty-three, twenty-three to twenty-five. Number of children: seven, six, nine, thirteen, f, five, four. Solution: Class marks xi: twelve, fourteen, sixteen, eighteen, twenty, twenty-two, twenty-four. Frequencies fi: seven, six, nine, thirteen, f, five, four. fi xi: eighty-four, eighty-four, one hundred forty-four, two hundred thirty-four, twenty f, one hundred ten, ninety-six. Sum of fi is forty-four plus f. Sum of fi xi is seven hundred fifty-two plus twenty f. Given mean is eighteen. So, eighteen equals seven hundred fifty-two plus twenty f divided by forty-four plus f. Eighteen times forty-four plus f equals seven hundred fifty-two plus twenty f. Seven hundred ninety-two plus eighteen f equals seven hundred fifty-two plus twenty f. Twenty f minus eighteen f equals seven hundred ninety-two minus seven hundred fifty-two. Two f equals forty. f equals twenty. The missing frequency is twenty.
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Question 4: Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method. Number of heartbeats per minute: sixty-five to sixty-eight, sixty-eight to seventy-one, seventy-one to seventy-four, seventy-four to seventy-seven, seventy-seven to eighty, eighty to eighty-three, eighty-three to eighty-six. Number of women: two, four, three, eight, seven, four, two. Solution: Class marks xi: sixty-six point five, sixty-nine point five, seventy-two point five, seventy-five point five, seventy-eight point five, eighty-one point five, eighty-four point five. Let a equals seventy-five point five, h equals three. Deviations di: minus nine, minus six, minus three, zero, three, six, nine. ui: minus three, minus two, minus one, zero, one, two, three. fi ui: minus six, minus eight, minus three, zero, seven, eight, six. Sum of fi is thirty. Sum of fi ui is four. Mean x bar equals a plus h times sigma fiui divided by sigma fi equals seventy-five point five plus three times four divided by thirty equals seventy-five point five plus zero point four equals seventy-five point nine. The mean heartbeats per minute is seventy-five point nine.
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Question 5: In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes. Number of mangoes: fifty to fifty-two, fifty-three to fifty-five, fifty-six to fifty-eight, fifty-nine to sixty-one, sixty-two to sixty-four. Number of boxes: fifteen, one hundred ten, one hundred thirty-five, one hundred fifteen, twenty-five. Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose? Solution: The classes are discontinuous. We adjust them by subtracting zero point five from lower limits and adding zero point five to upper limits. New classes: forty-nine point five to fifty-two point five, fifty-two point five to fifty-five point five, fifty-five point five to fifty-eight point five, fifty-eight point five to sixty-one point five, sixty-one point five to sixty-four point five. Class marks xi: fifty-one, fifty-four, fifty-seven, sixty, sixty-three. Let a equals fifty-seven, h equals three. Deviations di: minus six, minus three, zero, three, six. ui: minus two, minus one, zero, one, two. fi ui: minus thirty, minus one hundred ten, zero, one hundred fifteen, fifty. Sum of fi is four hundred. Sum of fi ui is minus twenty-five. Mean x bar equals fifty-seven plus three times minus twenty-five divided by four hundred equals fifty-seven minus zero point one eight seven five equals fifty-six point eight one two five. I used the step-deviation method because the class marks and frequencies are large.
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Question 6: The table below shows the daily expenditure on food of twenty-five households in a locality. Daily expenditure in rupees: one hundred to one hundred fifty, one hundred fifty to two hundred, two hundred to two hundred fifty, two hundred fifty to three hundred, three hundred to three hundred fifty. Number of households: four, five, twelve, two, two. Find the mean daily expenditure on food by a suitable method. Solution: Class marks xi: one hundred twenty-five, one hundred seventy-five, two hundred twenty-five, two hundred seventy-five, three hundred twenty-five. Let a equals two hundred twenty-five, h equals fifty. Deviations di: minus one hundred, minus fifty, zero, fifty, one hundred. ui: minus two, minus one, zero, one, two. fi ui: minus eight, minus five, zero, two, four. Sum of fi is twenty-five. Sum of fi ui is minus seven. Mean x bar equals two hundred twenty-five plus fifty times minus seven divided by twenty-five equals two hundred twenty-five minus fourteen equals two hundred eleven. The mean daily expenditure is two hundred eleven rupees.
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Question 7: To find out the concentration of SO2 in the air in parts per million, the data was collected for thirty localities in a certain city and is presented below. Concentration of SO2 in ppm: zero point zero zero to zero point zero four, zero point zero four to zero point zero eight, zero point zero eight to zero point twelve, zero point twelve to zero point sixteen, zero point sixteen to zero point twenty, zero point twenty to zero point twenty-four. Frequency: four, nine, nine, two, four, two. Find the mean concentration of SO2 in the air. Solution: Class marks xi: zero point zero two, zero point zero six, zero point ten, zero point fourteen, zero point eighteen, zero point twenty-two. fi xi: zero point zero eight, zero point five four, zero point nine zero, zero point two eight, zero point seven two, zero point four four. Sum of fi is thirty. Sum of fi xi is two point nine six. Mean x bar equals two point nine six divided by thirty equals zero point zero nine eight seven. The mean concentration is zero point zero nine nine ppm approximately.
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Question 8: A class teacher has the following absentee record of forty students of a class for the whole term. Find the mean number of days a student was absent. Number of days: zero to six, six to ten, ten to fourteen, fourteen to twenty, twenty to twenty-eight, twenty-eight to thirty-eight, thirty-eight to forty. Number of students: eleven, ten, seven, four, four, three, one. Solution: Class marks xi: three, eight, twelve, seventeen, twenty-four, thirty-three, thirty-nine. fi xi: thirty-three, eighty, eighty-four, sixty-eight, ninety-six, ninety-nine, thirty-nine. Sum of fi is forty. Sum of fi xi is four hundred ninety-nine. Mean x bar equals four hundred ninety-nine divided by forty equals twelve point four seven five. The mean number of days absent is twelve point four eight.
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Question 9: The following table gives the literacy rate in percentage of thirty-five cities. Find the mean literacy rate. Literacy rate in percent: forty-five to fifty-five, fifty-five to sixty-five, sixty-five to seventy-five, seventy-five to eighty-five, eighty-five to ninety-five. Number of cities: three, ten, eleven, eight, three. Solution: Class marks xi: fifty, sixty, seventy, eighty, ninety. Let a equals seventy, h equals ten. Deviations di: minus twenty, minus ten, zero, ten, twenty. ui: minus two, minus one, zero, one, two. fi ui: minus six, minus ten, zero, eight, six. Sum of fi is thirty-five. Sum of fi ui is minus two. Mean x bar equals seventy plus ten times minus two divided by thirty-five equals seventy minus zero point five seven one four equals sixty-nine point four three. The mean literacy rate is sixty-nine point four three percent.
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Now, let us move to Section 13.3, Mode of Grouped Data. Recall from Class Nine, a mode is that value among the observations which occurs most often, that is, the value of the observation having the maximum frequency. Further, we discussed finding the mode of ungrouped data. Here, we shall discuss ways of obtaining a mode of grouped data. It is possible that more than one value may have the same maximum frequency. In such situations, the data is said to be multimodal. Though grouped data can also be multimodal, we shall restrict ourselves to problems having a single mode only. Let us first recall how we found the mode for ungrouped data through the following example. Example 4: The wickets taken by a bowler in ten cricket matches are as follows: two, six, four, five, zero, two, one, three, two, three. Find the mode of the data. Solution: Let us form the frequency distribution table of the given data as follows: Number of wickets zero, one, two, three, four, five, six. Number of matches one, one, three, two, one, one, one. Clearly, two is the number of wickets taken by the bowler in the maximum number, that is, three, of matches. So, the mode of this data is two.
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In a grouped frequency distribution, it is not possible to determine the mode by looking at the frequencies. Here, we can only locate a class with the maximum frequency, called the modal class. The mode is a value inside the modal class, and is given by the formula: Mode equals l plus f1 minus f0 divided by two f1 minus f0 minus f2 times h, where l equals lower limit of the modal class, h equals size of the class interval assuming all class sizes to be equal, f1 equals frequency of the modal class, f0 equals frequency of the class preceding the modal class, f2 equals frequency of the class succeeding the modal class. Let us consider the following examples to illustrate the use of this formula. Example 5: A survey conducted on twenty households in a locality by a group of students resulted in the following frequency table for the number of family members in a household: Family size one to three, three to five, five to seven, seven to nine, nine to eleven. Number of families seven, eight, two, two, one. Find the mode of this data. Solution: Here the maximum class frequency is eight, and the class corresponding to this frequency is three to five. So, the modal class is three to five. Now modal class equals three to five, lower limit l of modal class equals three, class size h equals two, frequency f1 of the modal class equals eight, frequency f0 of class preceding the modal class equals seven, frequency f2 of class succeeding the modal class equals two. Now, let us substitute these values in the formula: Mode equals three plus eight minus seven divided by two times eight minus seven minus two times two equals three plus one divided by seven times two equals three plus zero point two eight six equals three point two eight six. Therefore, the mode of the data above is three point two eight six.
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Example 6: The marks distribution of thirty students in a mathematics examination are given in Table 13.3 of Example 1. Find the mode of this data. Also compare and interpret the mode and the mean. Solution: Refer to Table 13.3 of Example 1. Since the maximum number of students, that is, seven, have got marks in the interval forty to fifty-five, the modal class is forty to fifty-five. Therefore, the lower limit l of the modal class equals forty, the class size h equals fifteen, the frequency f1 of modal class equals seven, the frequency f0 of the class preceding the modal class equals three, the frequency f2 of the class succeeding the modal class equals six. Now, using the formula: Mode equals l plus f1 minus f0 divided by two f1 minus f0 minus f2 times h, we get Mode equals forty plus seven minus three divided by two times seven minus three minus six times fifteen equals forty plus four divided by five times fifteen equals forty plus twelve equals fifty-two. So, the mode marks is fifty-two. Now, from Example 1, you know that the mean marks is sixty-two. So, the maximum number of students obtained fifty-two marks, while on an average a student obtained sixty-two marks. Remarks: One: In Example 6, the mode is less than the mean. But for some other problems it may be equal or more than the mean also. Two: It depends upon the demand of the situation whether we are interested in finding the average marks obtained by the students or the average of the marks obtained by most of the students. In the first situation, the mean is required and in the second situation, the mode is required.
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Activity 3: Continuing with the same groups as formed in Activity 2 and the situations assigned to the groups. Ask each group to find the mode of the data. They should also compare this with the mean, and interpret the meaning of both. Remark: The mode can also be calculated for grouped data with unequal class sizes. However, we shall not be discussing it. Let us move to Exercise 13.2. Question 1: The following table shows the ages of the patients admitted in a hospital during a year: Age in years: five to fifteen, fifteen to twenty-five, twenty-five to thirty-five, thirty-five to forty-five, forty-five to fifty-five, fifty-five to sixty-five. Number of patients: six, eleven, twenty-one, twenty-three, fourteen, five. Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency. Solution: For mode: Modal class is thirty-five to forty-five. l equals thirty-five, h equals ten, f1 equals twenty-three, f0 equals twenty-one, f2 equals fourteen. Mode equals thirty-five plus twenty-three minus twenty-one divided by two times twenty-three minus twenty-one minus fourteen times ten equals thirty-five plus two divided by eleven times ten equals thirty-five plus one point eight two equals thirty-six point eight two. For mean: Class marks xi: ten, twenty, thirty, forty, fifty, sixty. fi xi: sixty, two hundred twenty, six hundred thirty, nine hundred twenty, seven hundred, three hundred. Sum fi is eighty. Sum fi xi is two thousand eight hundred thirty. Mean x bar equals two thousand eight hundred thirty divided by eighty equals thirty-five point three eight. The mode is thirty-six point eight two and mean is thirty-five point three eight. The maximum number of patients are in the age group around thirty-six point eight two, while the average age is thirty-five point three eight.
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Question 2: The following data gives the information on the observed lifetimes in hours of two hundred twenty-five electrical components: Lifetimes: zero to twenty, twenty to forty, forty to sixty, sixty to eighty, eighty to one hundred, one hundred to one hundred twenty. Frequency: ten, thirty-five, fifty-two, sixty-one, thirty-eight, twenty-nine. Determine the modal lifetimes of the components. Solution: Modal class is sixty to eighty. l equals sixty, h equals twenty, f1 equals sixty-one, f0 equals fifty-two, f2 equals thirty-eight. Mode equals sixty plus sixty-one minus fifty-two divided by two times sixty-one minus fifty-two minus thirty-eight times twenty equals sixty plus nine divided by thirty-two times twenty equals sixty plus five point six two five equals sixty-five point six two five. The modal lifetime is sixty-five point six three hours.
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Question 3: The following data gives the distribution of total monthly household expenditure of two hundred families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure. Expenditure in rupees: one thousand to one thousand five hundred, one thousand five hundred to two thousand, two thousand to two thousand five hundred, two thousand five hundred to three thousand, three thousand to three thousand five hundred, three thousand five hundred to four thousand, four thousand to four thousand five hundred, four thousand five hundred to five thousand. Number of families: twenty-four, forty, thirty-three, twenty-eight, thirty, twenty-two, sixteen, seven. Solution: For mode: Modal class is one thousand five hundred to two thousand. l equals one thousand five hundred, h equals five hundred, f1 equals forty, f0 equals twenty-four, f2 equals thirty-three. Mode equals one thousand five hundred plus forty minus twenty-four divided by two times forty minus twenty-four minus thirty-three times five hundred equals one thousand five hundred plus sixteen divided by twenty-three times five hundred equals one thousand five hundred plus three hundred forty-seven point eight three equals one thousand eight hundred forty-seven point eight three. For mean: Class marks xi: one thousand two hundred fifty, one thousand seven hundred fifty, two thousand two hundred fifty, two thousand seven hundred fifty, three thousand two hundred fifty, three thousand seven hundred fifty, four thousand two hundred fifty, four thousand seven hundred fifty. Let a equals two thousand seven hundred fifty, h equals five hundred. Deviations di: minus one thousand five hundred, minus one thousand, minus five hundred, zero, five hundred, one thousand, one thousand five hundred, two thousand. ui: minus three, minus two, minus one, zero, one, two, three, four. fi ui: minus seventy-two, minus eighty, minus thirty-three, zero, thirty, forty-four, forty-eight, twenty-eight. Sum fi is two hundred. Sum fi ui is minus thirty-five. Mean x bar equals two thousand seven hundred fifty plus five hundred times minus thirty-five divided by two hundred equals two thousand seven hundred fifty minus eighty-seven point five equals two thousand six hundred sixty-two point five. The modal expenditure is one thousand eight hundred forty-seven point eight three rupees and mean is two thousand six hundred sixty-two point five rupees.
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Question 4: The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures. Number of students per teacher: fifteen to twenty, twenty to twenty-five, twenty-five to thirty, thirty to thirty-five, thirty-five to forty, forty to forty-five, forty-five to fifty, fifty to fifty-five. Number of states or U T: three, eight, nine, ten, three, zero, zero, two. Solution: Modal class is thirty to thirty-five. l equals thirty, h equals five, f1 equals ten, f0 equals nine, f2 equals three. Mode equals thirty plus ten minus nine divided by two times ten minus nine minus three times five equals thirty plus one divided by eight times five equals thirty plus zero point six two five equals thirty point six three. For mean: Class marks xi: seventeen point five, twenty-two point five, twenty-seven point five, thirty-two point five, thirty-seven point five, forty-two point five, forty-seven point five, fifty-two point five. fi xi: fifty-two point five, one hundred eighty, two hundred forty-seven point five, three hundred twenty-five, one hundred twelve point five, zero, zero, one hundred five. Sum fi is thirty-five. Sum fi xi is nine hundred twenty-two point five. Mean x bar equals nine hundred twenty-two point five divided by thirty-five equals twenty-six point three six. The mode thirty point six three indicates most states have around thirty students per teacher, while the mean twenty-six point three six is the average ratio.
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Question 5: The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches. Runs scored: three thousand to four thousand, four thousand to five thousand, five thousand to six thousand, six thousand to seven thousand, seven thousand to eight thousand, eight thousand to nine thousand, nine thousand to ten thousand, ten thousand to eleven thousand. Number of batsmen: four, eighteen, nine, seven, six, three, one, one. Find the mode of the data. Solution: Modal class is four thousand to five thousand. l equals four thousand, h equals one thousand, f1 equals eighteen, f0 equals four, f2 equals nine. Mode equals four thousand plus eighteen minus four divided by two times eighteen minus four minus nine times one thousand equals four thousand plus fourteen divided by twenty-three times one thousand equals four thousand plus six hundred eight point seven equals four thousand six hundred eight point seven. The mode is four thousand six hundred eight point seven runs.
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Question 6: A student noted the number of cars passing through a spot on a road for one hundred periods each of three minutes and summarised it in the table given below. Find the mode of the data: Number of cars: zero to ten, ten to twenty, twenty to thirty, thirty to forty, forty to fifty, fifty to sixty, sixty to seventy, seventy to eighty. Frequency: seven, fourteen, thirteen, twelve, twenty, eleven, fifteen, eight. Solution: Modal class is forty to fifty. l equals forty, h equals ten, f1 equals twenty, f0 equals twelve, f2 equals eleven. Mode equals forty plus twenty minus twelve divided by two times twenty minus twelve minus eleven times ten equals forty plus eight divided by seventeen times ten equals forty plus four point seven one equals forty-four point seven one. The mode is forty-four point seven one cars.
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Now, let us move to Section 13.4, Median of Grouped Data. As you have studied in Class Nine, the median is a measure of central tendency which gives the value of the middle-most observation in the data. Recall that for finding the median of ungrouped data, we first arrange the data values of the observations in ascending order. Then, if n is odd, the median is the n plus one divided by two th observation. And, if n is even, then the median will be the average of the n divided by two th and the n divided by two plus one th observations. Suppose, we have to find the median of the following data, which gives the marks, out of fifty, obtained by one hundred students in a test: Marks obtained twenty, twenty-nine, twenty-eight, thirty-three, forty-two, thirty-eight, forty-three, twenty-five. Number of students six, twenty-eight, twenty-four, fifteen, two, four, one, twenty. First, we arrange the marks in ascending order and prepare a frequency table as follows: Marks twenty, twenty-five, twenty-eight, twenty-nine, thirty-three, thirty-eight, forty-two, forty-three. Frequencies six, twenty, twenty-four, twenty-eight, fifteen, four, two, one. Total one hundred. Here n equals one hundred, which is even. The median will be the average of the n divided by two th and the n divided by two plus one th observations, that is, the fiftieth and fifty-first observations. To find these observations, we proceed as follows: We add another column depicting cumulative frequency. Marks twenty has cumulative frequency six. Marks twenty-five has twenty-six. Marks twenty-eight has fifty. Marks twenty-nine has seventy-eight. Marks thirty-three has ninety-three. Marks thirty-eight has ninety-seven. Marks forty-two has ninety-nine. Marks forty-three has one hundred.
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From the table above, we see that: fiftieth observation is twenty-eight. Fifty-first observation is twenty-nine. So, Median equals twenty-eight plus twenty-nine divided by two equals twenty-eight point five. Remark: The part of the table consisting the marks column and the cumulative frequency column is known as Cumulative Frequency Table. The median marks twenty-eight point five conveys the information that about fifty percent students obtained marks less than twenty-eight point five and another fifty percent students obtained marks more than twenty-eight point five. Now, let us see how to obtain the median of grouped data, through the following situation. Consider a grouped frequency distribution of marks obtained, out of one hundred, by fifty-three students, in a certain examination, as follows: Marks zero to ten, ten to twenty, twenty to thirty, thirty to forty, forty to fifty, fifty to sixty, sixty to seventy, seventy to eighty, eighty to ninety, ninety to one hundred. Number of students five, three, four, three, three, four, seven, nine, seven, eight. From the table above, try to answer the following questions: How many students have scored marks less than ten? The answer is clearly five. How many students have scored less than twenty marks? Observe that the number of students who have scored less than twenty include the number of students who have scored marks from zero to ten as well as the number of students who have scored marks from ten to twenty. So, the total number of students with marks less than twenty is five plus three, that is, eight. We say that the cumulative frequency of the class ten to twenty is eight.
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Similarly, we can compute the cumulative frequencies of the other classes, that is, the number of students with marks less than thirty, less than forty, dot dot dot, less than one hundred. We give them in a table: Less than ten: five. Less than twenty: eight. Less than thirty: twelve. Less than forty: fifteen. Less than fifty: eighteen. Less than sixty: twenty-two. Less than seventy: twenty-nine. Less than eighty: thirty-eight. Less than ninety: forty-five. Less than one hundred: fifty-three. The distribution given above is called the cumulative frequency distribution of the less than type. Here ten, twenty, thirty, dot dot dot, one hundred, are the upper limits of the respective class intervals. We can similarly make the table for the number of students with scores, more than or equal to zero, more than or equal to ten, more than or equal to twenty, and so on. From the frequency table, we observe that all fifty-three students have scored marks more than or equal to zero. Since there are five students scoring marks in the interval zero to ten, this means that there are fifty-three minus five equals forty-eight students getting more than or equal to ten marks. Continuing in the same manner, we get the number of students scoring twenty or above as forty-eight minus three equals forty-five, thirty or above as forty-five minus four equals forty-one, and so on. The table is: More than or equal to zero: fifty-three. More than or equal to ten: forty-eight. More than or equal to twenty: forty-five. More than or equal to thirty: forty-one. More than or equal to forty: thirty-eight. More than or equal to fifty: thirty-five. More than or equal to sixty: thirty-one. More than or equal to seventy: twenty-four. More than or equal to eighty: fifteen. More than or equal to ninety: eight. The table above is called a cumulative frequency distribution of the more than type. Here zero, ten, twenty, dot dot dot, ninety give the lower limits of the respective class intervals.
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Now, to find the median of grouped data, we can make use of any of these cumulative frequency distributions. Let us combine the frequency and cumulative frequency tables to get a combined table. Marks zero to ten, ten to twenty, twenty to thirty, thirty to forty, forty to fifty, fifty to sixty, sixty to seventy, seventy to eighty, eighty to ninety, ninety to one hundred. Frequencies five, three, four, three, three, four, seven, nine, seven, eight. Cumulative frequencies five, eight, twelve, fifteen, eighteen, twenty-two, twenty-nine, thirty-eight, forty-five, fifty-three. Now in a grouped data, we may not be able to find the middle observation by looking at the cumulative frequencies as the middle observation will be some value in a class interval. It is, therefore, necessary to find the value inside a class that divides the whole distribution into two halves. But which class should this be? To find this class, we find the cumulative frequencies of all the classes and n divided by two. We now locate the class whose cumulative frequency is greater than and nearest to n divided by two. This is called the median class. In the distribution above, n equals fifty-three. So, n divided by two equals twenty-six point five. Now sixty to seventy is the class whose cumulative frequency twenty-nine is greater than and nearest to twenty-six point five. Therefore, sixty to seventy is the median class. After finding the median class, we use the following formula for calculating the median. Median equals l plus n divided by two minus cf divided by f times h, where l equals lower limit of median class, n equals number of observations, cf equals cumulative frequency of class preceding the median class, f equals frequency of median class, h equals class size assuming class size to be equal.
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Substituting the values n divided by two equals twenty-six point five, l equals sixty, cf equals twenty-two, f equals seven, h equals ten in the formula above, we get Median equals sixty plus twenty-six point five minus twenty-two divided by seven times ten equals sixty plus four point five divided by seven times ten equals sixty plus six point four three equals sixty-six point four three. So, about half the students have scored marks less than sixty-six point four three, and the other half have scored marks more than sixty-six point four three. Example 7: A survey regarding the heights in centimetres of fifty-one girls of Class Ten of a school was conducted and the following data was obtained: Height in cm: Less than one hundred forty, Less than one hundred forty-five, Less than one hundred fifty, Less than one hundred fifty-five, Less than one hundred sixty, Less than one hundred sixty-five. Number of girls: four, eleven, twenty-nine, forty, forty-six, fifty-one. Find the median height. Solution: To calculate the median height, we need to find the class intervals and their corresponding frequencies. The given distribution being of the less than type, one hundred forty, one hundred forty-five, one hundred fifty, dot dot dot, one hundred sixty-five give the upper limits of the corresponding class intervals. So, the classes should be below one hundred forty, one hundred forty to one hundred forty-five, one hundred forty-five to one hundred fifty, dot dot dot, one hundred sixty to one hundred sixty-five. Observe that from the given distribution, we find that there are four girls with height less than one hundred forty, that is, the frequency of class interval below one hundred forty is four. Now, there are eleven girls with heights less than one hundred forty-five and four girls with height less than one hundred forty. Therefore, the number of girls with height in the interval one hundred forty to one hundred forty-five is eleven minus four equals seven. Similarly, the frequency of one hundred forty-five to one hundred fifty is twenty-nine minus eleven equals eighteen, for one hundred fifty to one hundred fifty-five, it is forty minus twenty-nine equals eleven, for one hundred fifty-five to one hundred sixty, it is forty-six minus forty equals six, for one hundred sixty to one hundred sixty-five, it is fifty-one minus forty-six equals five.
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So, our frequency distribution table with the given cumulative frequencies becomes: Class intervals: Below one hundred forty, one hundred forty to one hundred forty-five, one hundred forty-five to one hundred fifty, one hundred fifty to one hundred fifty-five, one hundred fifty-five to one hundred sixty, one hundred sixty to one hundred sixty-five. Frequencies: four, seven, eighteen, eleven, six, five. Cumulative frequencies: four, eleven, twenty-nine, forty, forty-six, fifty-one. Now n equals fifty-one. So, n divided by two equals twenty-five point five. This observation lies in the class one hundred forty-five to one hundred fifty. Then, l the lower limit equals one hundred forty-five, cf the cumulative frequency of the class preceding one hundred forty-five to one hundred fifty equals eleven, f the frequency of the median class one hundred forty-five to one hundred fifty equals eighteen, h the class size equals five. Using the formula, Median equals l plus n divided by two minus cf divided by f times h, we have Median equals one hundred forty-five plus twenty-five point five minus eleven divided by eighteen times five equals one hundred forty-five plus fourteen point five divided by eighteen times five equals one hundred forty-five plus four point zero three equals one hundred forty-nine point zero three. So, the median height of the girls is one hundred forty-nine point zero three cm. This means that the height of about fifty percent of the girls is less than this height, and fifty percent are taller than this height.
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Example 8: The median of the following data is five hundred twenty-five. Find the values of x and y, if the total frequency is one hundred. Class intervals: zero to one hundred, one hundred to two hundred, two hundred to three hundred, three hundred to four hundred, four hundred to five hundred, five hundred to six hundred, six hundred to seven hundred, seven hundred to eight hundred, eight hundred to nine hundred, nine hundred to one thousand. Frequencies: two, five, x, twelve, seventeen, twenty, y, nine, seven, four. Solution: We form a cumulative frequency table. Cumulative frequencies: two, seven, seven plus x, nineteen plus x, thirty-six plus x, fifty-six plus x, fifty-six plus x plus y, sixty-five plus x plus y, seventy-two plus x plus y, seventy-six plus x plus y. It is given that n equals one hundred. So, seventy-six plus x plus y equals one hundred, that is, x plus y equals twenty-four. Equation one. The median is five hundred twenty-five, which lies in the class five hundred to six hundred. So, l equals five hundred, f equals twenty, cf equals thirty-six plus x, h equals one hundred. Using the formula: Median equals l plus n divided by two minus cf divided by f times h, we get five hundred twenty-five equals five hundred plus fifty minus thirty-six minus x divided by twenty times one hundred. That is, five hundred twenty-five minus five hundred equals fourteen minus x times five. That is, twenty-five equals seventy minus five x. That is, five x equals seventy minus twenty-five equals forty-five. So, x equals nine. Therefore, from equation one, we get nine plus y equals twenty-four, that is, y equals fifteen.
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Now, that you have studied about all the three measures of central tendency, let us discuss which measure would be best suited for a particular requirement. The mean is the most frequently used measure of central tendency because it takes into account all the observations, and lies between the extremes, that is, the largest and the smallest observations of the entire data. It also enables us to compare two or more distributions. For example, by comparing the average mean results of students of different schools of a particular examination, we can conclude which school has a better performance. However, extreme values in the data affect the mean. For example, the mean of classes having frequencies more or less the same is a good representative of the data. But, if one class has frequency, say two, and the five others have frequency twenty, twenty-five, twenty, twenty-one, eighteen, then the mean will certainly not reflect the way the data behaves. So, in such cases, the mean is not a good representative of the data. In problems where individual observations are not important, and we wish to find out a typical observation, the median is more appropriate, for example, finding the typical productivity rate of workers, average wage in a country, and so on. These are situations where extreme values may be there. So, rather than the mean, we take the median as a better measure of central tendency. In situations which require establishing the most frequent value or most popular item, the mode is the best choice, for example, to find the most popular television programme being watched, the consumer item in greatest demand, the colour of the vehicle used by most of the people, and so on.
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Remarks: One: There is an empirical relationship between the three measures of central tendency: three Median equals Mode plus two Mean. Two: The median of grouped data with unequal class sizes can also be calculated. However, we shall not discuss it here. Let us move to Exercise 13.3. Question 1: The following frequency distribution gives the monthly consumption of electricity of sixty-eight consumers of a locality. Find the median, mean and mode of the data and compare them. Monthly consumption in units: sixty-five to eighty-five, eighty-five to one hundred five, one hundred five to one hundred twenty-five, one hundred twenty-five to one hundred forty-five, one hundred forty-five to one hundred sixty-five, one hundred sixty-five to one hundred eighty-five, one hundred eighty-five to two hundred five. Number of consumers: four, five, thirteen, twenty, fourteen, eight, four. Solution: For median: n equals sixty-eight, n divided by two equals thirty-four. Cumulative frequencies: four, nine, twenty-two, forty-two, fifty-six, sixty-four, sixty-eight. Median class is one hundred twenty-five to one hundred forty-five. l equals one hundred twenty-five, cf equals twenty-two, f equals twenty, h equals twenty. Median equals one hundred twenty-five plus thirty-four minus twenty-two divided by twenty times twenty equals one hundred twenty-five plus twelve equals one hundred thirty-seven. For mean: Class marks xi: seventy-five, ninety-five, one hundred fifteen, one hundred thirty-five, one hundred fifty-five, one hundred seventy-five, one hundred ninety-five. Let a equals one hundred thirty-five, h equals twenty. Deviations di: minus sixty, minus forty, minus twenty, zero, twenty, forty, sixty. ui: minus three, minus two, minus one, zero, one, two, three. fi ui: minus twelve, minus ten, minus thirteen, zero, fourteen, sixteen, twelve. Sum fi is sixty-eight. Sum fi ui is seven. Mean x bar equals one hundred thirty-five plus twenty times seven divided by sixty-eight equals one hundred thirty-five plus two point zero six equals one hundred thirty-seven point zero six. For mode: Modal class is one hundred twenty-five to one hundred forty-five. l equals one hundred twenty-five, f1 equals twenty, f0 equals thirteen, f2 equals fourteen, h equals twenty. Mode equals one hundred twenty-five plus twenty minus thirteen divided by two times twenty minus thirteen minus fourteen times twenty equals one hundred twenty-five plus seven divided by thirteen times twenty equals one hundred twenty-five plus ten point seven seven equals one hundred thirty-five point seven seven. The median, mean, and mode are very close, indicating a fairly symmetric distribution.
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Question 2: If the median of the distribution given below is twenty-eight point five, find the values of x and y. Class interval: zero to ten, ten to twenty, twenty to thirty, thirty to forty, forty to fifty, fifty to sixty. Frequency: five, x, twenty, fifteen, y, five. Total sixty. Solution: Cumulative frequencies: five, five plus x, twenty-five plus x, forty plus x, forty plus x plus y, forty-five plus x plus y. Total n equals sixty. So, forty-five plus x plus y equals sixty, x plus y equals fifteen. Equation one. Median is twenty-eight point five, which lies in twenty to thirty. l equals twenty, f equals twenty, cf equals five plus x, h equals ten. Median equals twenty plus thirty minus five minus x divided by twenty times ten equals twenty plus twenty-five minus x divided by two equals twenty-eight point five. So, twenty-five minus x divided by two equals eight point five. Twenty-five minus x equals seventeen. x equals eight. From equation one, eight plus y equals fifteen, so y equals seven.
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Question 3: A life insurance agent found the following data for distribution of ages of one hundred policy holders. Calculate the median age, if policies are given only to persons having age eighteen years onwards but less than sixty year. Age in years: Below twenty, Below twenty-five, Below thirty, Below thirty-five, Below forty, Below forty-five, Below fifty, Below fifty-five, Below sixty. Number of policy holders: two, six, twenty-four, forty-five, seventy-eight, eighty-nine, ninety-two, ninety-eight, one hundred. Solution: Convert to continuous classes: eighteen to twenty, twenty to twenty-five, twenty-five to thirty, thirty to thirty-five, thirty-five to forty, forty to forty-five, forty-five to fifty, fifty to fifty-five, fifty-five to sixty. Frequencies: two, four, eighteen, twenty-one, thirty-three, eleven, three, six, two. Cumulative frequencies: two, six, twenty-four, forty-five, seventy-eight, eighty-nine, ninety-two, ninety-eight, one hundred. n equals one hundred, n divided by two equals fifty. Median class is thirty-five to forty. l equals thirty-five, cf equals forty-five, f equals thirty-three, h equals five. Median equals thirty-five plus fifty minus forty-five divided by thirty-three times five equals thirty-five plus five divided by thirty-three times five equals thirty-five plus zero point seven six equals thirty-five point seven six. The median age is thirty-five point seven six years.
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Question 4: The lengths of forty leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table: Length in mm: one hundred eighteen to one hundred twenty-six, one hundred twenty-seven to one hundred thirty-five, one hundred thirty-six to one hundred forty-four, one hundred forty-five to one hundred fifty-three, one hundred fifty-four to one hundred sixty-two, one hundred sixty-three to one hundred seventy-one, one hundred seventy-two to one hundred eighty. Number of leaves: three, five, nine, twelve, five, four, two. Find the median length of the leaves. Hint: The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to one hundred seventeen point five to one hundred twenty-six point five, one hundred twenty-six point five to one hundred thirty-five point five, dot dot dot, one hundred seventy-one point five to one hundred eighty point five. Solution: Adjusted classes: one hundred seventeen point five to one hundred twenty-six point five, one hundred twenty-six point five to one hundred thirty-five point five, one hundred thirty-five point five to one hundred forty-four point five, one hundred forty-four point five to one hundred fifty-three point five, one hundred fifty-three point five to one hundred sixty-two point five, one hundred sixty-two point five to one hundred seventy-one point five, one hundred seventy-one point five to one hundred eighty point five. Frequencies: three, five, nine, twelve, five, four, two. Cumulative frequencies: three, eight, seventeen, twenty-nine, thirty-four, thirty-eight, forty. n equals forty, n divided by two equals twenty. Median class is one hundred forty-four point five to one hundred fifty-three point five. l equals one hundred forty-four point five, cf equals seventeen, f equals twelve, h equals nine. Median equals one hundred forty-four point five plus twenty minus seventeen divided by twelve times nine equals one hundred forty-four point five plus two point two five equals one hundred forty-six point seven five. The median length is one hundred forty-six point seven five mm.
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Question 5: The following table gives the distribution of the life time of four hundred neon lamps: Life time in hours: one thousand five hundred to two thousand, two thousand to two thousand five hundred, two thousand five hundred to three thousand, three thousand to three thousand five hundred, three thousand five hundred to four thousand, four thousand to four thousand five hundred, four thousand five hundred to five thousand. Number of lamps: fourteen, fifty-six, sixty, eighty-six, seventy-four, sixty-two, forty-eight. Find the median life time of a lamp. Solution: Cumulative frequencies: fourteen, seventy, one hundred thirty, two hundred sixteen, two hundred ninety, three hundred fifty-two, four hundred. n equals four hundred, n divided by two equals two hundred. Median class is three thousand to three thousand five hundred. l equals three thousand, cf equals one hundred thirty, f equals eighty-six, h equals five hundred. Median equals three thousand plus two hundred minus one hundred thirty divided by eighty-six times five hundred equals three thousand plus three hundred forty-eight point eight four equals three thousand three hundred forty-eight point eight four. The median life time is three thousand three hundred forty-eight point eight four hours.
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Question 6: One hundred surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of letters: one to four, four to seven, seven to ten, ten to thirteen, thirteen to sixteen, sixteen to nineteen. Number of surnames: six, thirty, forty, sixteen, four, four. Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames. Solution: For median: Cumulative frequencies: six, thirty-six, seventy-six, ninety-two, ninety-six, one hundred. n divided by two equals fifty. Median class is four to seven. l equals four, cf equals six, f equals thirty, h equals three. Median equals four plus fifty minus six divided by thirty times three equals four plus four point four equals eight point four. For mean: Class marks xi: two point five, five point five, eight point five, eleven point five, fourteen point five, seventeen point five. fi xi: fifteen, one hundred sixty-five, three hundred forty, one hundred eighty-four, fifty-eight, seventy. Sum fi is one hundred. Sum fi xi is eight hundred thirty-two. Mean x bar equals eight hundred thirty-two divided by one hundred equals eight point three two. For mode: Modal class is seven to ten. l equals seven, f1 equals forty, f0 equals thirty, f2 equals sixteen, h equals three. Mode equals seven plus forty minus thirty divided by two times forty minus thirty minus sixteen times three equals seven plus ten divided by twenty-eight times three equals seven plus one point zero seven equals eight point zero seven. Median is eight point four, mean is eight point three two, mode is eight point zero seven.
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Question 7: The distribution below gives the weights of thirty students of a class. Find the median weight of the students. Weight in kg: forty to forty-five, forty-five to fifty, fifty to fifty-five, fifty-five to sixty, sixty to sixty-five, sixty-five to seventy, seventy to seventy-five. Number of students: two, three, eight, six, six, three, two. Solution: Cumulative frequencies: two, five, thirteen, nineteen, twenty-five, twenty-eight, thirty. n equals thirty, n divided by two equals fifteen. Median class is fifty-five to sixty. l equals fifty-five, cf equals thirteen, f equals six, h equals five. Median equals fifty-five plus fifteen minus thirteen divided by six times five equals fifty-five plus one point six seven equals fifty-six point six seven. The median weight is fifty-six point six seven kg.
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Now, let us conclude with Section 13.5, Summary. In this chapter, you have studied the following points: One: The mean for grouped data can be found by the direct method: x bar equals sigma fi xi divided by sigma fi. Two: The assumed mean method: x bar equals a plus sigma fi di divided by sigma fi. Three: The step deviation method: x bar equals a plus h times sigma fi ui divided by sigma fi, with the assumption that the frequency of a class is centred at its mid-point, called its class mark. Four: The mode for grouped data can be found by using the formula: Mode equals l plus f1 minus f0 divided by two f1 minus f0 minus f2 times h, where symbols have their usual meanings. Five: The cumulative frequency of a class is the frequency obtained by adding the frequencies of all the classes preceding the given class. Six: The median for grouped data is formed by using the formula: Median equals l plus n divided by two minus cf divided by f times h, where symbols have their usual meanings. A Note to the Reader: For calculating mode and median for grouped data, it should be ensured that the class intervals are continuous before applying the formulae.
Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]