A class master maintains the following record of absence of students in his class. Number of students absent Number of days 0 2 1 7 2 6 3 8 4 30 5 25 6 7 7 12 8 8 9 5 Total = 110 Find the means number of students absent per day.

A class master maintains the following record of absence of students in his class. Number of students absent x Number of days f f x 0 2 0 1 7 7 2 6 12 3 8 24 4 30 120 5 25 125 6 7 42 7 12 84 8 8 64 9 5 45 Σ f = 110 Σ f x = 523 We have to find the means number of students absent per day. Mean = Σ f x Σ f Mean number of students absent per day = 523 110 = 4 . 754 Hence, the required answer is 4 . 75 .

The expenditure for the consumption of water per month by 100 families is given below. Expenditure on water (in ₹) Number of families 30 - 40 12 40 - 50 18 50 - 60 20 60 - 70 15 70 - 80 12 80 - 90 11 90 - 100 6 100 - 110 4 110 - 120 2 Find the mean monthly expenditure of the families on water. Also, determine the quartiles of the expenditure.

The expenditure for the consumption of water per month by 100 families is given below. We have to find the mean monthly expenditure of the families on water and also, determine the quartiles of the expenditure. Expenditure on water (in ₹) Class interval x Number of families f f x 30 - 40 35 12 420 40 - 50 45 18 810 50 - 60 55 20 1100 60 - 70 65 15 975 70 - 80 75 12 900 80 - 90 85 11 935 90 - 100 95 6 570 100 - 110 105 4 420 110 - 120 115 2 230 Σ f = 100 Σ f x = 6360 Mean of the observation = Σ f x Σ f Mean monthly expenditure of the families on water = Σ f x Σ f = 6360 100 = 63 . 6 The first or lower Quartile Q 1 = l + N 4 - c f × h First Quartile class is 40 - 50 . Q 1 = 40 + 100 4 - 12 18 × 10 = 40 + 13 18 × 10 = 47 . 22 Third or upper Quartile Q 3 = l + 3 4 N - c f × h Third Quartile class is 60 - 70 . Q 3 = 60 + 3 4 × 100 - 50 15 × 10 = 60 + 75 - 50 15 × 10 Q 3 = 60 + 25 15 × 10 = 60 + 16 . 66 = 76 . 66

Find the median, lower and upper quartiles, 4 t h decile and 56 t h percentile of the following distribution of marks: Marks Number of students 0 - 4 10 4 - 8 12 8 - 12 18 12 - 14 7 14 - 18 5 18 - 20 8 20 - 25 4 25 and above 6

We have to find the median, lower and upper quartiles, 4 t h decile and 50 t h percentile of the following distribution of marks: Marks Number of students Cumulative frequency 0 - 4 10 10 4 - 8 12 22 8 - 12 18 40 12 - 14 7 47 14 - 18 5 52 18 - 20 8 60 20 - 25 4 64 25 and above 6 70 Here, N is 70 so N 2 = 70 2 = 35 Median class = 8 - 12 Median = M = 8 + a fraction of the class width 4 We assume that the frequency 18 of the median class is uniformly distributed over the width i.e, 4 of the class. We use the formula, M = l + N 2 - c f × h where, l = lower limit of the median class = 8 N = size of the sample or population = 70 c = cumulative frequency of the class just before the median class = 22 f = frequency of the median class = 18 h = width of the median class = 4 ⇒ M = 8 + 70 2 - 22 18 × 4 = 8 + 35 - 22 18 × 4 ⇒ M = 8 + 13 9 × 2 = 10 . 89 The first or lower Quartile Q 1 = l + N 4 - c f × h First Quartile class is 4 - 8 Q 1 = 4 + 70 4 - 10 12 × 4 = 4 + 17 . 5 - 10 12 × 4 = 4 + 0 . 625 × 4 = 6 . 5 Third or upper Quartile Q 3 = l + 3 4 N - c f × h Third Quartile class is 12 - 14 . Q 3 = 14 + 3 4 × 70 - 47 5 × 4 = 14 + 52 . 5 - 47 5 × 4 Q 3 = 14 + 5 . 5 5 × 4 = 14 + 4 . 4 = 18 . 4 In same way we have to find 4 t h decile or the 40 t h percentile of the data given by formula, D 4 = P 40 = l + 40 100 N - c f × h ⇒ 40 100 × 70 = 28 , so class is 8 - 12 D 4 = 8 + 40 100 × 70 - 22 18 × 4 = 8 + 28 - 22 18 × 4 = 8 + 6 18 × 4 = 8 + 4 3 = 9 . 33 P 56 = l + 56 100 N - c f × h ⇒ 56 100 × 70 = 39 . 2 , so class is 8 - 12 P 56 = 8 + 39 . 2 - 22 18 × 4 = 8 + 17 . 2 18 × 4 P 56 = 8 + 0 . 96 × 4 = 11 . 82

Find the median, the two quartiles and the mode of the following distribution. Marks below Number of students 10 15 20 35 30 60 40 84 50 96 60 127 70 198 80 250

We have to find the median, the two quartiles and the mode of the following distribution. For finding median we need to write the class interval and the frequency. Marks below Class interval Number of students (frequency) Cumulative frequency 10 0 - 10 15 15 20 10 - 20 20 35 30 20 - 30 25 60 40 30 - 40 24 84 50 40 - 50 12 96 60 50 - 60 31 127 70 60 - 70 71 198 80 70 - 80 52 250 N = 250 , N 2 = 250 2 = 125 Thus, cumulative frequency near to N 2 is 127 , so the median class is 50 - 60 . We use the formula, M = l + N 2 - c f × h where, l = lower limit of the median class N = size of the sample or population c = cumulative frequency of the class just before the median class f = frequency of the median class h = width of the median class l = 50 , N = 250 , c = 96 , f = 31 , h = 10 ⇒ M = 50 + 250 2 - 96 31 × 10 = 50 + 125 - 96 31 × 10 = 50 + 29 31 × 10 = 50 + 142 . 5 127 × 10 = 50 + 9 . 35 = 59 . 35 Hence, the median is 59 . 35 . The first or lower Quartile Q 1 = l + N 4 - c f × h ⇒ N 4 = 250 4 = 62 . 5 , so first Quartile class is 19 - 29 . l = 20 , N = 250 , c = 35 , f = 25 , h = 10 Q 1 = 20 + 250 4 - 35 25 × 10 = 20 + 62 . 5 - 35 25 × 10 = 20 + 27 . 5 25 × 10 = 20 + 11 = 30 Third or upper Quartile Q 3 = l + 3 4 N - c f × h ⇒ 3 4 N = 3 × 250 4 = 187 . 5 Third Quartile class is 60 - 70 . Q 3 = 60 + 3 4 × 250 - 127 71 × 10 = 60 + 187 . 5 - 127 198 × 10 Q 3 = 60 + 60 . 5 71 × 10 = 60 + 8 . 52 = 68 . 52 From the table we can state that 60 - 70 is the modal class as it has the highest frequency. For finding the mode, we use the formula Mode = l + f m - f 1 2 f m - f 1 - f 2 × h ⇒ 60 + 71 - 31 2 × 71 - 31 - 52 × 10 = 60 + 40 59 × 10 = 60 + 6 . 78 = 66 . 78

Draw the less than ogive graph and indicate the positions of the quartiles. Marks below Number of students 10 15 20 35 30 60 40 84 50 96 60 127 70 198 80 250

For drawing the less than ogive graph and locating the quartiles we need to write the class interval and the frequency. Marks below Class interval Number of students (frequency) Cumulative frequency 10 0 - 10 15 15 20 10 - 20 20 35 30 20 - 30 25 60 40 30 - 40 24 84 50 40 - 50 12 96 60 50 - 60 31 127 70 60 - 70 71 198 80 70 - 80 52 250 For a grouped frequency distribution, the less than graph is drawn by marking points with the upper limit of each class as the abscissa and the corresponding less than cumulative frequency of the class as ordinate. These points are joined by free hand. Now we have to plot the points 10 , 15 20 , 35 30 , 60 40 , 84 50 , 96 60 , 127 70 , 198 80 , 250 Here we have from the table N = 250 , N 2 = 250 2 = 125 , N 4 = 250 4 = 62 . 5 , 3 N 4 = 187 . 5 Along the y - a x i s , we locate points corresponding to coordinates N 2 = 125 , N 4 = 62 . 5 , 3 N 4 = 187 . 5 Through these points, lines are drawn parallel to the x - a x i s meeting the ogive at A , B , and C respectively. The abscissae of these points give Q 1 , Q 2 , Q 3 respectively and they are such that Q 1 > 30 , Q 2 < 60 , Q 3 < 70 .

Find the median, lower and upper quartiles, 8 t h decile, 56 t h percentile and the mode of the following distribution of 245 workers. Daily savings (in ₹) Number of workers 1 - 2 . 99 6 3 - 4 . 99 53 5 - 6 . 99 85 7 - 8 . 99 56 9 - 10 . 99 21 11 - 12 . 99 16 13 - 14 . 99 4 15 - 16 . 99 4

Daily savings (in ₹) Class interval Number of workers Cumulative frequency 1 - 2 . 99 1 - 2 . 995 6 6 3 - 4 . 99 2 . 995 - 4 . 995 53 59 5 - 6 . 99 4 . 995 - 6 . 995 85 144 7 - 8 . 99 6 . 995 - 8 . 995 56 200 9 - 10 . 99 8 . 995 - 10 . 995 21 221 11 - 12 . 99 10 . 995 - 12 . 995 16 237 13 - 14 . 99 12 . 995 - 14 . 995 4 241 15 - 16 . 99 14 . 995 - 16 . 995 4 245 N = 245 , N 2 = 245 2 = 122 . 5 Thus, cumulative frequency near to N 2 is 144 , so the median class is 4 . 995 - 6 . 995 . We use the formula, M = l + N 2 - c f × h where, l = lower limit of the median class N = size of the sample or population c = cumulative frequency of the class just before the median class f = frequency of the median class h = width of the median class l = 4 . 995 , N = 245 , c = 59 , f = 85 , h = 2 ⇒ M = 4 . 995 + 245 2 - 59 85 × 2 = 4 . 995 + 122 . 5 - 59 85 × 2 = 4 . 995 + 63 . 5 85 × 2 = 4 . 995 + 0 . 747 × 2 = 4 . 995 + 1 . 494 = 6 . 49 Hence, the median is 6 . 49 . The first or lower Quartile Q 1 = l + N 4 - c f × h ⇒ N 4 = 245 4 = 61 . 25 , so first Quartile class is 2 . 995 - 4 . 995 . l = 2 . 995 , N = 245 , c = 6 , f = 53 , h = 2 Q 1 = 2 . 995 + 245 4 - 6 53 × 2 = 2 . 995 + 61 . 25 - 6 53 × 2 = 2 . 995 + 55 . 25 53 × 2 = 2 . 995 + 2 . 085 = 5 . 08 Third or upper Quartile Q 3 = l + 3 4 N - c f × h ⇒ 3 4 N = 3 × 245 4 = 183 . 75 Third Quartile class is 6 . 995 - 8 . 995 . Q 3 = 6 . 995 + 3 4 × 245 - 144 56 × 2 = 6 . 995 + 183 . 75 - 144 56 × 2 Q 3 = 6 . 995 + 39 . 75 56 × 2 = 6 . 995 + 1 . 4196 = 8 . 41 In same way we have to find 8 t h decile or the 80 t h percentile of the data given by formula, D 8 = P 80 = l + 80 100 N - c f × h ⇒ 80 100 × 245 = 196 , so class is 6 . 995 - 8 . 995 D 8 = 6 . 995 + 80 100 × 245 - 144 56 × 2 = 6 . 995 + 196 - 144 56 × 2 = 6 . 995 + 52 56 × 2 = 6 . 995 + 1 . 9259 = 8 . 92 P 56 = l + 56 100 N - c f × h ⇒ 56 100 × 245 = 137 . 2 , so class is 4 . 995 - 6 . 995 P 56 = 4 . 995 + 137 . 2 - 59 85 × 2 = 4 . 995 + 78 . 2 85 × 2 P 56 = 4 . 995 + 78 . 2 85 × 2 = 4 . 995 + 1 . 84 = 6 . 84 From the table we can state that 4 . 995 - 6 . 995 is the modal class as it has the highest frequency. For finding the mode, we use the formula Mode = l + f m - f 1 2 f m - f 1 - f 2 × h ⇒ 4 . 995 + 85 - 53 2 × 85 - 53 - 56 × 2 = 4 . 995 + 32 61 × 2 = 4 . 995 + 1 . 0491 = 6 . 04 Hence, mode is 6 . 04 .

Manipur Board Solutions for Chapter: Statistics, Exercise 1: EXERCISE 13.1

Author:Manipur Board

Manipur Board Mathematics Solutions for Exercise - Manipur Board Solutions for Chapter: Statistics, Exercise 1: EXERCISE 13.1

Attempt the practice questions on Chapter 13: Statistics, Exercise 1: EXERCISE 13.1 with hints and solutions to strengthen your understanding. Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from Manipur Board Solutions for Chapter: Statistics, Exercise 1: EXERCISE 13.1 with Hints & Solutions

MEDIUM

10th Manipur Board

IMPORTANT

Mathematics for Class 10>Chapter 13 - Statistics>EXERCISE 13.1>Q 5

A class master maintains the following record of absence of students in his class.

Number of students absent	Number of days
$0$	$2$
$1$	$7$
$2$	$6$
$3$	$8$
$4$	$30$
$5$	$25$
$6$	$7$
$7$	$12$
$8$	$8$
$9$	$5$
	Total $= 110$

Find the means number of students absent per day.

HARD

10th Manipur Board

IMPORTANT

Mathematics for Class 10>Chapter 13 - Statistics>EXERCISE 13.1>Q 6

The following is the frequency distribution of the number of teachers in Higher Secondary Schools in $1978$ in India. Find the average number of teachers per Higher Secondary School in India for $1978$ .

Number of Teachers	Number of Higher Secondary Schools
$6 - 10$	$955$
$11 - 15$	$1067$
$16 - 20$	$1663$
$21 - 25$	$1492$
$26 - 30$	$1220$
$31 - 35$	$1129$
$36 - 40$	$745$
$41 - 45$	$637$
$46 - 50$	$442$

HARD

10th Manipur Board

IMPORTANT

Mathematics for Class 10>Chapter 13 - Statistics>EXERCISE 13.1>Q 7

The expenditure for the consumption of water per month by $100$ families is given below.

Expenditure on water (in ₹)	Number of families
$30 - 40$	$12$
$40 - 50$	$18$
$50 - 60$	$20$
$60 - 70$	$15$
$70 - 80$	$12$
$80 - 90$	$11$
$90 - 100$	$6$
$100 - 110$	$4$
$110 - 120$	$2$

Find the mean monthly expenditure of the families on water. Also, determine the quartiles of the expenditure.

HARD

10th Manipur Board

IMPORTANT

Mathematics for Class 10>Chapter 13 - Statistics>EXERCISE 13.1>Q 8

Find the median, lower and upper quartiles, $4 t h$ decile and $56 t h$ percentile of the following distribution of marks:

Marks	Number of students
$0 - 4$	$10$
$4 - 8$	$12$
$8 - 12$	$18$
$12 - 14$	$7$
$14 - 18$	$5$
$18 - 20$	$8$
$20 - 25$	$4$
$25$ and above	$6$

MEDIUM

10th Manipur Board

IMPORTANT

Mathematics for Class 10>Chapter 13 - Statistics>EXERCISE 13.1>Q 9

A school has $4$ sections in class $X$ having $45, 50, 42$ and $30$ students respectively. A test in Mathematics was conducted in all the sections and the mean marks of these sections are $62, 55, 58$ and $49$ respectively. Find the overall average mark of the class.

HARD

10th Manipur Board

IMPORTANT

Mathematics for Class 10>Chapter 13 - Statistics>EXERCISE 13.1>Q 11

Find the median, the two quartiles and the mode of the following distribution.

Marks below	Number of students
$10$	$15$
$20$	$35$
$30$	$60$
$40$	$84$
$50$	$96$
$60$	$127$
$70$	$198$
$80$	$250$

HARD

10th Manipur Board

IMPORTANT

Mathematics for Class 10>Chapter 13 - Statistics>EXERCISE 13.1>Q 11

Draw the less than ogive graph and indicate the positions of the quartiles.

Marks below	Number of students
$10$	$15$
$20$	$35$
$30$	$60$
$40$	$84$
$50$	$96$
$60$	$127$
$70$	$198$
$80$	$250$

HARD

10th Manipur Board

IMPORTANT

Mathematics for Class 10>Chapter 13 - Statistics>EXERCISE 13.1>Q 12

Find the median, lower and upper quartiles, $8 t h$ decile, $56 t h$ percentile and the mode of the following distribution of $245$ workers.

Daily savings (in ₹)	Number of workers
$1 - 2.99$	$6$
$3 - 4.99$	$53$
$5 - 6.99$	$85$
$7 - 8.99$	$56$
$9 - 10.99$	$21$
$11 - 12.99$	$16$
$13 - 14.99$	$4$
$15 - 16.99$	$4$

Manipur Board Solutions for Chapter: Statistics, Exercise 1: EXERCISE 13.1

Manipur Board Mathematics Solutions for Exercise - Manipur Board Solutions for Chapter: Statistics, Exercise 1: EXERCISE 13.1

Questions from Manipur Board Solutions for Chapter: Statistics, Exercise 1: EXERCISE 13.1 with Hints & Solutions

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