Welcome dear students! Today we are going to learn about Data Handling from Class 7 Maths. You might be aware of the term average and would have come across statements involving the term average in your day to day life. Isha spends on an average of about 5 hours daily for her studies. The average temperature at this time of the year is about 40 degree celsius. The average age of pupils in my class is 12 years. The average attendance of students in a school during its final examination was 98 per cent. Think about these statements. Do you think the child studies exactly 5 hours daily? Or is the temperature always 40 degrees? Or is every pupil exactly 12 years old? Obviously not. By average we understand that Isha usually studies for 5 hours. On some days less, on others more. The average temperature of 40 degree celsius means very often the temperature is around 40. Sometimes less, sometimes more. Thus, average is a number that represents or shows the central tendency of a group of observations or data. Since average lies between the highest and the lowest value of the given data, we say average is a measure of the central tendency of the group of data. Different forms of data need different forms of representative value. One of these is the Arithmetic mean. You will learn about others later.
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Now let us move to section 3.2, Arithmetic Mean. The most common representative value is the arithmetic mean or simply the mean. To understand this, consider two vessels containing 20 litres and 60 litres of milk. What is the amount each vessel would have if they share equally? We are seeking the arithmetic mean. The average would be (20 + 60) / 2 litres = 40 litres. Thus, each vessel has 40 litres. The mean is defined as: mean = Sum of all observations / number of observations. Let us see examples. Example 1: Ashish studies 4 hours, 5 hours and 3 hours on three consecutive days. Average study time = (4 + 5 + 3) / 3 hours = 12 / 3 hours = 4 hours per day. Ashish studies 4 hours daily on average. Example 2: A batsman scored 36, 35, 50, 46, 60, 55 runs in six innings. Total runs = 36 + 35 + 50 + 46 + 60 + 55 = 282. Mean = 282 / 6 = 47. The mean runs per inning are 47.
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Where does the arithmetic mean lie? Try These: How would you find the average of your study hours for the whole week? Add all seven days and divide by 7. Think, Discuss and Write: Consider the data in the examples. Is the mean bigger than each observation? Is it smaller than each observation? Discuss with your friends. You will find the mean lies between the greatest and smallest observations. The mean of two numbers always lies between them. For example, mean of 5 and 11 is (5 + 11) / 2 = 8, which lies between 5 and 11. Now, I want you to frame one more example of this type and answer the same questions. Can you use this to find fractional numbers between 1/2 and 1/4? Yes, their average is (1/2 + 1/4) / 2 = 3/8. Between 1/2 and 3/8, the average is 7/16, and so on. Try These: 1. Find the mean of your sleeping hours during one week. 2. Find at least 5 numbers between 1/2 and 1/3 by repeatedly taking averages.
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Next is section 3.2.1, Range. The difference between the highest and lowest observation gives the spread. This is the range. Example 3: Ages of 10 teachers are 32, 41, 28, 54, 35, 26, 23, 33, 38, 40. First, arrange in ascending order: 23, 26, 28, 32, 33, 35, 38, 40, 41, 54. The oldest teacher is 54 years, and the youngest is 23 years. Second, the range equals 54 minus 23, which is 31 years. Third, the mean age equals the sum of all ages, which is 350, divided by 10. The mean age is 35 years.
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Now, Exercise 3.1. Question 1: Find the range of heights of any ten students. Measure them, find tallest and shortest, subtract. Question 2: Organise marks: 4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7. Highest is 9. Lowest is 1. Range equals 9 minus 1, which is 8. Mean equals sum divided by 20, which is 100 divided by 20, giving 5. Question 3: Mean of first five whole numbers (0, 1, 2, 3, 4). Sum is 10. Mean equals 10 divided by 5, which is 2. Question 4: Cricketer runs: 58, 76, 40, 35, 46, 45, 0, 100. Sum is 400. Mean equals 400 divided by 8, which is 50. Question 5: Player table. A's mean equals (14 + 16 + 10 + 10) divided by 4, which is 12.5. For C, divide total by 3 because C did not play in game 3. B's mean equals (0 + 8 + 6 + 4) divided by 4, which is 4.5. C's mean equals (8 + 11 + 13) divided by 3, which is 10.67. A is the best performer.
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Question 6: Science test marks: 85, 76, 90, 85, 39, 48, 56, 95, 81, 75. Highest is 95, Lowest is 39. Range equals 95 minus 39, which is 56. Mean equals sum 730 divided by 10, which is 73. Question 7: Enrolment: 1555, 1670, 1750, 2013, 2540, 2820. Sum is 12348. Mean equals 12348 divided by 6, which is 2058. Question 8: Rainfall in mm: Mon 0.0, Tue 12.2, Wed 2.1, Thu 0.0, Fri 20.5, Sat 5.5, Sun 1.0. Range equals 20.5 minus 0.0, which is 20.5. Mean equals sum 41.3 divided by 7, which is 5.9. Days with rainfall less than 5.9 are Monday, Wednesday, Thursday, Saturday, and Sunday. That is 5 days. Question 9: Heights of 10 girls: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141. Tallest is 151 cm. Shortest is 128 cm. Range equals 151 minus 128, which is 23 cm. Sum is 1414. Mean equals 1414 divided by 10, which is 141.4 cm. Girls taller than 141.4 cm are 150, 151, 146, 149, and 143. That is 5 girls.
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Now section 3.3, Mode. Mean is not the only measure of central tendency. Look at this example: A shopkeeper records shirt sales for sizes 90, 95, 100, 105, 110 cm. Numbers sold: 8, 22, 32, 37, 6. Total is 105. Mean equals 105 divided by 5, which is 21. Should he stock 21 of each size? No. He decides to stock 95, 100, 105 cm sizes because they sell most. Another example: A dress shop owner says the most popular size is 90 cm. She looks at the size sold the most. This is the mode. The mode of a set of observations is the observation that occurs most often. Example 4: Find the mode of 1, 1, 2, 4, 3, 2, 1, 2, 2, 4. Arrange: 1, 1, 1, 2, 2, 2, 2, 3, 4, 4. Mode is 2 because it occurs most frequently.
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Section 3.3.1: Mode of Large Data. For large data, we tabulate. Example 5: Football victory margins: 1, 3, 2, 5, 1, 4, 6, 2, 5, 2, 2, 2, 4, 1, 2, 3, 1, 1, 2, 3, 2, 6, 4, 3, 2, 1, 1, 4, 2, 1, 5, 3, 3, 2, 3, 2, 4, 2, 1, 2. Tabulating frequencies: 1 appears 9 times. 2 appears 14 times. 3 appears 7 times. 4 appears 5 times. 5 appears 3 times. 6 appears 2 times. Total 40 matches. Mode is 2, as it occurs most. Most matches won by a margin of 2 goals. Now, let us solve the Try These on Page 61 explicitly. Question (i): Find the mode of 2, 6, 5, 3, 0, 3, 4, 3, 2, 4, 5, 2, 4. Counting frequencies: 0 appears once, 2 appears three times, 3 appears three times, 4 appears three times, 5 appears twice, 6 appears once. Here, 2, 3, and 4 all occur the maximum number of times. So, the data has three modes: 2, 3, and 4. Question (ii): Find the mode of 2, 14, 16, 12, 14, 14, 16, 14, 10, 14, 18, 14. Counting frequencies: 14 appears six times, which is the highest. So the mode is 14.
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Think, Discuss and Write: Can a set of numbers have more than one mode? Yes. Example 6: Numbers: 2, 2, 2, 3, 3, 4, 5, 5, 5, 6, 6, 8. Here 2 and 5 both occur 3 times. Both are modes. Try These: 1. Record classmate ages, tabulate, find mode. 2. Record heights, find mode. Now, let us look at the Try These on Page 62. 1. Find mode of: 12, 14, 12, 16, 15, 13, 14, 18, 19, 12, 14, 15, 16, 15, 16, 16, 15, 17, 13, 16, 16, 15, 15, 13, 15, 17, 15, 14, 15, 13, 15, 14. Counting: 12 appears 3 times. 13 appears 4 times. 14 appears 5 times. 15 appears 10 times. 16 appears 7 times. 17 appears 2 times. 18 appears 1 time. 19 appears 1 time. Mode is 15. 2. Heights of 25 children: 168, 165, 163, 160, 163, 161, 162, 164, 163, 162, 164, 163, 160, 163, 160, 165, 163, 162, 163, 164, 163, 160, 165, 163, 162. Counting: 163 appears 9 times. Mode is 163.
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Now, let us do the Do This activity on Page 62. We will analyze four situations to decide which representative value is appropriate. Situation (a): You have to decide the number of chapattis needed for 25 people. Suppose the number of chapattis each person needs is: 2, 3, 2, 3, 2, 1, 2, 3, 2, 2, 4, 2, 2, 3, 2, 4, 4, 2, 3, 2, 4, 2, 4, 3, 5. The mode is 2 chapattis. If we use mode, we prepare 2 times 25 equals 50 chapattis. This is clearly inadequate for the group. Would mean be appropriate? Yes, calculating the mean gives a better estimate for total food needed. Situation (b): A shopkeeper selling shirts decides to replenish her stock. She should look at which size sells the most. Here, the mode is the best representative value to decide stock. Situation (c): Finding the height of a door for a house. Suppose 5 children are around 135 cm and 4 adults are taller. The mode is 135 cm. If we make a door based on mode, it will be too short for adults. Mode is inappropriate. Mean is also not ideal. We should use the maximum height to decide the door height. Situation (d): Going on a picnic, if only one fruit can be bought for everyone. We should buy the fruit most people prefer. Here, mode is the appropriate representative value. Now, discuss with your friends and frame two situations where mean is appropriate, and two situations where mode is appropriate.
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Now section 3.4, Median. Sometimes mean or mode fails. Consider 17 students' heights: 106, 110, 123, 125, 117, 120, 112, 115, 110, 120, 115, 102, 115, 115, 109, 115, 101. To divide into two equal groups by height, mean is 1930 divided by 17, which is 113.5. Groups would be 7 and 10 members. Not equal. Mode is 115. Groups would be 7 below, 10 at or above. Not equal. So we use Median. Arrange in ascending order: 101, 102, 106, 109, 110, 110, 112, 115, 115, 115, 115, 115, 117, 120, 120, 123, 125. The middle value is 115. It divides the data into 8 above and 8 below. This is the Median. Median is the value in the middle when arranged in order, with half above and half below. We consider odd observations here. Mean, mode, and median are all measures of central tendency. Example 7: Find median of 24, 36, 46, 17, 18, 25, 35. Arrange: 17, 18, 24, 25, 35, 36, 46. Middle is 25. Median is 25.
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Exercise 3.2. Question 1: Scores: 19, 25, 23, 20, 9, 20, 15, 10, 5, 16, 25, 20, 24, 12, 20. Arrange: 5, 9, 10, 12, 15, 16, 19, 20, 20, 20, 20, 23, 24, 25, 25. Median is the 8th value, which is 20. Mode is 20. They are same. Question 2: Runs: 6, 15, 120, 50, 100, 80, 10, 15, 8, 10, 15. Arrange: 6, 8, 10, 10, 15, 15, 15, 50, 80, 100, 120. Median is the 6th value, which is 15. Mode is 15. Mean equals 429 divided by 11, which is 39. Not same. Question 3: Friend's data: 35, 32, 35, 42, 38, 32, 34. Friend said Median equals 42, Mode equals 32. Correct: Arrange: 32, 32, 34, 35, 35, 38, 42. Median is 35. Mode is 32 and 35. Friend was wrong. Question 4: Weights: 38, 42, 35, 37, 45, 50, 32, 43, 43, 40, 36, 38, 43, 38, 47. Arrange: 32, 35, 36, 37, 38, 38, 38, 40, 42, 43, 43, 43, 45, 47, 50. Median is the 8th value, which is 40. Mode is 38 and 43. Yes, more than one mode. Question 5: True or False. (i) Mode is always in data. True. (ii) Mean is always in data. False. (iii) Median is always in data. True for odd number of observations. (iv) Data 6, 4, 3, 8, 9, 12, 13, 9 has mean 9. Sum is 64. Mean equals 64 divided by 8, which is 8. False.
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Section 3.5: Use of Bar Graphs. We arrange data in tables and show it visually. You can deduce information. If bars represent frequency, the longest bar shows the mode. 3.5.1 Choosing a Scale. Bar lengths depend on frequency and scale. One unit can represent 1, 10, or 100 observations. Example 8: 200 students chose favourite colours. Red 43, Green 19, Blue 55, Yellow 49, Orange 34. To graph: Scale 0 to 60, 1 unit equals 10 students. Draw vertical bars for each colour. Conclusions: Blue is most preferred because the bar is tallest. Green is least preferred because the bar is shortest. Five colours total. Example 9: Marks out of 600. Ajay 450, Bali 500, Dipti 300, Faiyaz 360, Geetika 400, Hari 540. Scale: 1 unit equals 100 marks. Draw bars accordingly.
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Double Bar Graphs compare two data sets side by side. Consider average daily sunshine hours. Margate: Jan 2, Feb 4, Mar 4, Apr 8, May 7, Jun 6, Jul 4, Aug 2, Sep 3, Oct 1, Nov 4, Dec 7. Aberdeen: Jan 1/2, Feb 3/4, Mar 1, Apr 1/2, May 3, Jun 4, Jul 7, Aug 1/2, Sep 3, Oct 1/2, Nov 1/2, Dec 1/4. Draw two bars per month. Except April, Margate has more sunshine. Example 10: Teacher checks new technique. Quarterly vs Half-yearly scores for 5 students. Ashish 10 to 15. Arun 15 to 18. Kavish 12 to 16. Maya 20 to 21. Rita 9 to 15. Double bar graph shows marked improvement. Teacher continues technique. Now, let us fully solve the Try These after Example 10. First, the water resistant watches survey. The bar graph shows the number of watches tested and the number that leaked for different companies. To find the fraction of watches that leaked, divide the number of leaked watches by the total number tested for each company. For instance, if Company A had 20 leaks out of 50 tested, the fraction is 20 over 50. Calculate this for every company. The company with the smallest fraction has the best quality watches because fewer leaked relative to the number tested. Second, the book sales data. In 1995, English sold 350 and Hindi sold 500. In 1996, English 400, Hindi 525. In 1997, English 450, Hindi 600. In 1998, English 620, Hindi 650. Draw a double bar graph with years on the horizontal axis and sales on the vertical axis. Question (a): Find the difference each year. 1995: 150. 1996: 125. 1997: 150. 1998: 30. The least difference was in 1998. Question (b): Did English demand rise faster? Yes. English rose from 350 to 620, an increase of 270. Hindi rose from 500 to 650, an increase of 150. English sales grew at a faster rate.
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Exercise 3.3. Question 1: Pet bar graph. Looking at the provided graph, the tallest bar corresponds to the cat, so it is the most popular pet. The dog bar reaches the 8 mark, meaning 8 students have a dog as a pet. Question 2: Book sales graph. 1989 approximately 180, 1990 approximately 475, 1992 approximately 225. 475 sold in 1990. 225 sold in 1992. Fewer than 250 in 1989 and 1992. Estimate 1989 by checking bar height below 200 mark. Question 3: Class sizes. Fifth 135, Sixth 120, Seventh 95, Eighth 100, Ninth 90, Tenth 80. Scale 0 to 140, increments of 20. Max: Fifth. Min: Tenth. Ratio Sixth to Eighth equals 120 to 100, which simplifies to 6:5. Question 4: Term scores. English 67 to 70. Hindi 72 to 65. Maths 88 to 95. Science 81 to 85. S. Science 73 to 75. Most improvement: Maths with plus 7. Least: S. Science with plus 2. Down: Hindi with minus 7.
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Question 5: Colony sports survey. Watching: Cricket 1240, Basketball 470, Swimming 510, Hockey 430, Athletics 250. Participating: 620, 320, 320, 250, 105. Draw double bar graph. Watching bars are taller for all sports. Most popular: Cricket. Watching is more preferred. Question 6: City temperatures. Referring to the temperature table provided at the start of the chapter, we have the following minimum and maximum temperatures: Ahmedabad min 27, max 38. Amritsar min 26, max 37. Bangalore min 21, max 28. Chennai min 25, max 32. Delhi min 26, max 39. Jaipur min 29, max 42. Jammu min 26, max 41. Mumbai min 27, max 32. Now, plot a double bar graph and answer: Largest difference in min and max? Jammu has 41 minus 26 equals 15, which is the largest. Hottest city is Jaipur at 42. Coldest is Bangalore at 21. Two cities where max of one is less than min of another: Bangalore's maximum of 28 is less than Jaipur's minimum of 29. Least difference: Mumbai, with 32 minus 27 equals 5.
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What have we discussed? 1. Average represents central tendency. 2. Arithmetic mean is a representative value. 3. Mode is the most frequent observation. 4. Median is the middle value in ordered data. 5. Bar graphs use uniform width bars. 6. Double bar graphs compare two data sets. I hope this lesson on Data Handling was clear and helpful. Practice calculating mean, mode, and median. Draw bar graphs to visualize your data. You are doing wonderfully! Keep practicing and you will master this chapter. Thank you for listening! Keep revising and practicing. Goodbye! [CHAPTER_COMPLETE]