## Chapter 3 Data and Handling Exercise 3.1

**Question 1: Find the range of heights of any ten students of your class.Answer:**

133, 143, 132, 211, 167, 110, 151, 199, 187, 201

The highest value is = 211 cm

The lowest value is = 110 cm

Range of heights = highest value - lowest value

= 211 - 110

= 101 cm

**Question 2: Organise the following marks in a class assessment, in a tabular form.**

4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7

(i) Which number is the highest?

(ii) Which number is the lowest?

(iii) What is the range of the data?

(iv) Find the arithmetic mean.

Answer:

4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7

(i) Which number is the highest?

(ii) Which number is the lowest?

(iii) What is the range of the data?

(iv) Find the arithmetic mean.

Answer:

(i) 9 is the highest number.

(ii) 1 is the lowest number.

(iii) Range of marks = highest value - lowest value

(iv) Arithmetic mean = sum of observation/number of observations

= 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9/9

= 45/9

= 5**Question 3: Find the mean of the first five whole numbers.Answer:** First five whole numbers = 0, 1, 2, 3, 4

Mean = sum of observations/number of observations

= 0 + 1 + 2 + 3 + 4/5

= 10/5

= 2

Therefore, the mean of first five whole numbers is 2.

**Question 4: A cricketer scores the following runs in eight innings:**

58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.

Answer:

58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.

Answer:

Mean = sum of observations/number of observations

= 58 + 76 + 40 + 35 + 46 + 45 + 0 + 100/8

= 400/8

= 50

Therefore, the mean score of runs in eight innings is 50.

**Question 5: Following table shows the points of each player scored in four games:**

Now answer the following questions:

(i) Find the mean to determine A’s average number of points scored per game.

(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?

(iii) B played in all the four games. How would you find the mean?

(iv) Who is the best performer?

Answer:

Now answer the following questions:

(i) Find the mean to determine A’s average number of points scored per game.

(ii) To find the mean number of points per game for C, would you divide the total points by 3 or by 4? Why?

(iii) B played in all the four games. How would you find the mean?

(iv) Who is the best performer?

Answer:

= 14 + 16 + 10 + 10/4

= 50/4

= 12.5

(ii) To find the mean number of points per game for C, we will divide the total points by 3 because C played only 3 games.

(iii) B played all the four games, so we will divide the total points by 4.

Mean = sum of observations/number of observations

= 0 + 8 + 6 + 4/4

= 18/4

= 4.5

(iv) Mean of Player A = sum of observations/number of observations

= 14 + 16 + 10 + 10/4

= 50/4

= 12.5

Mean of Player B = sum of observations/number of observations

= 0 + 8 + 6 + 4/4

= 18/4

= 4.5

Mean of Player C = sum of observations/number of observations

= 8 + 11 + 13/3

= 32/3

= 10.67

Compare

Player A mean = 12.5

Player B mean = 4.5

Player C mean = 10.67

In comparison, Player A has the highest mean.

Therefore, Player A is the best performer.

**Question 6: The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Find the:**

(i) Highest and the lowest marks obtained by the students.

(ii) Range of the marks obtained.

(iii) Mean marks obtained by the group.

Answer:

(i) Highest and the lowest marks obtained by the students.

(ii) Range of the marks obtained.

(iii) Mean marks obtained by the group.

Answer:

Ascending order of marks = 39, 48, 56, 75, 76, 81, 85, 85, 90, 95

(i) Highest marks = 95

Lowest marks = 39

(ii) Range = Highest marks - Lowest marks

= 95 - 39

= 56

(iii) Mean = sum of observations/number of observations

= 39, 48, 56, 75, 76, 81, 85, 85, 90, 95/10

= 730/10

= 73

**Question 7: The enrolment in a school during six consecutive years was as follows: 1555, 1670, 1750, 2013, 2540, 2820. Find the mean enrolment of the school for this period.**

Answer:

Answer:

Mean = Sum of all observations/ Number of observations

= 1555 + 1670 + 1750 + 2013 + 2540 + 2820/ 6

= 12348/6

= 2058

Therefore, the mean enrolment of the school for this given period is 2058.

**Question 8: The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:**

**(i) Find the range of the rainfall in the above data.**

(ii) Find the mean rainfall for the week.

(iii) On how many days was the rainfall less than the mean rainfall.

Answer:

(ii) Find the mean rainfall for the week.

(iii) On how many days was the rainfall less than the mean rainfall.

Answer:

(i) Range = Highest observation - Lowest observation

= 20.5 - 0.0

= 20.5 mm

(ii) Mean = Sum of observations/ Number of observation

= 0.0 + 12.2 + 2.1 + 0.0 + 20.5 + 5.5 + 1.0/ 7

= 41.3/7

= 5.9 mm

(iii) 5 (Monday, Wednesday, Thursday, Saturday and Sunday) days are the days in which the rainfall was less than the mean rainfall.

**Question 9: The heights of 10 girls were measured in cm and the results are as follows:**

135, 150, 139, 128, 151, 132, 146, 149, 143, 141.

(i) What is the height of the tallest girl?

(ii) What is the height of the shortest girl?

(iii) What is the range of the data?

(iv) What is the mean height of the girls?

(v) How many girls have heights more than the mean height.

Answer:

135, 150, 139, 128, 151, 132, 146, 149, 143, 141.

(i) What is the height of the tallest girl?

(ii) What is the height of the shortest girl?

(iii) What is the range of the data?

(iv) What is the mean height of the girls?

(v) How many girls have heights more than the mean height.

Answer:

Height of students in ascending order - 128, 132, 135, 139, 141, 143, 146, 149, 150, 151

(i) The height of the tallest girl is 151 cm.

(ii) The height of the shortest girl is 128 cm.

(iii) Range = Highest observation - Lowest observation

= 151 - 128

= 23 cm

(iv) Mean = Sum of observations/ Number of observations

= 128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151/ 10

= 1414/10

= 141.4 cm

(v) 5 girls have heights more than the mean height.

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