Find the mean of the following data using the assumed mean method: Class Frequency 0 - 5 8 5 - 10 7 10 - 15 10 15 - 20 13 20 - 25 12

Let us first construct the table for d i × f i , where d i = x i - A (Assumed mean), as shown below : Class Classmark x i Frequency d i = x i - A d i × f i 0 - 5 2 . 5 8 - 10 - 80 5 - 10 7 . 5 7 - 5 - 35 10 - 15 12 . 5 = A 10 0 0 15 - 20 17 . 5 13 5 65 20 - 25 22 . 5 12 10 120 Σ f i = 50 Σ f i d i = 70 Let A = 12 . 5 . Then, Σ f i d i = 70 . Now, the required mean, ( x ¯ ) = A + Σ f i d i Σ f i = 12 . 5 + 70 50 = 12 . 5 + 1 . 4 = 13 . 9 Therefore, the required mean is 13 . 9

The monthly expenditure on milk in 200 families of Housing Society is given below: The monthly expenditure in ₹ 1000 - 1500 1500 - 2000 2000 - 2500 2500 - 3000 3000 - 3500 3500 - 4000 4000 - 4500 4500 - 5000 The number of families 24 40 33 x 30 22 16 7 Find the value of x and also, find the median and mean expenditure on milk.

Given, The total number of families = 200 ⇒ 24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 200 x = 28 Expenditure (in ₹ ) No. of families f i Mid value x i x i × f i C.F. 1000 - 1500 24 1250 30000 24 1500 - 2000 40 1750 70000 64 2000 - 2500 33 2250 74250 97 2500 - 3000 28 2750 77000 125 3000 - 3500 30 3250 97500 155 3500 - 4000 22 3750 82500 177 4000 - 4500 16 4250 68000 193 4500 - 5000 7 4750 33250 200 ∑ f i = 200 ∑ f i x i = 532500 Mean = ∑ f i x i ∑ f i = 532500 200 = ₹ 2662 . 5 N 2 = 200 2 = 100 t h 100 t h observation lies in class 2500 - 3000 which is known as median class. l = lower limit of median class = 2500 N = number of families = 200 f = frequency of median class = 28 C F = cumulative frequency of the class preceding the median class = 97 h = class size = 500 Median = l + N 2 - C F f × h = 2500 + 200 2 - 97 28 × 500 = 2500 + 100 - 97 28 × 500 = 2500 + 3 28 × 500 = 2500 + 53 . 571 = 2553 . 571 Hence, this is the required answer.

The monthly expenditure on milk in 200 families of a housing society is given below: Monthly Expenditure(in ₹ ) 1000 - 1500 1500 - 2000 2000 - 2500 2500 - 3000 3000 - 3500 3500 - 4000 4000 - 4500 4500 - 5000 Number of families 24 40 33 x 30 22 16 7 Find the value of x , and also find the median and mean expenditure on milk.

From the given, total number of families = 200 ⇒ 24 + 40 + 33 + x + 30 + 22 + 16 + 7 = 200 ⇒ x = 200 - 172 = 28 Monthly expenditure (in ₹ ) (C.I) Number of families ( f i ) Mid value ( x i ) x i f i C.F 1000 - 1500 24 1250 30000 24 1500 - 2000 40 1750 70000 64 2000 - 2500 33 2250 74250 97 2500 - 3000 28 2750 77000 125 3000 - 3500 30 3250 97500 155 3500 - 4000 22 3750 82500 177 4000 - 4500 16 4250 68000 193 4500 - 5000 7 4750 33250 200 N = ∑ f i = 200 ∑ x i f i = 532500 Mean = ∑ x i f i ∑ f i = 532500 200 = ₹ 2662 . 5 Now, N 2 = 200 2 = 100 th 100 th observation lies in class 2500 - 3000 which lies is known as median class. Lower limit of median l = 2500 Number of families N = 200 Frequency of median class f = 28 Cumulative frequency of the class preceding the median class C . F = 97 Class size h = 500 Median = l + N 2 - C . F f × h Median = 2500 + 200 2 - 97 28 × 500 ⇒ Median = 2500 + 100 - 97 28 × 500 ⇒ Median = 2500 + 3 7 × 125 ⇒ Median = 2500 + 53 . 571 ⇒ Median = 2553 . 571 Hence, the value of x is 28 , mean and median is ₹ 2662 . 5 and 2553 . 571 respectively.

250 apples of a box were weighed and the distribution of masses of the apples is given in the following table: Mass (in grams) 80 - 100 100 - 120 120 - 140 140 - 160 160 - 180 Number of apples 20 60 70 x 60 Find the value of x , the mean mass of the apples and the modal mass of the apples.

Given, total number of apples = 250 20 + 60 + 70 + x + 60 = 250 x = 250 - 210 = 40 Mass (in grams) Number of apples ( f ) Mid value ( x ) d = x i - A f × d 80 - 100 20 90 - 40 - 800 100 - 120 60 110 - 20 - 1200 120 - 140 70 130 A 0 0 140 - 160 40 150 20 800 160 - 180 60 170 40 2400 ∑ f = 250 ∑ f d = 1200 Now, mean x ¯ = A + ∑ f d ∑ f = 130 + 1200 250 = 134 . 8 Now, Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h Where / = lower class limit of modal class = 12 Modal class is ( 120 - 140 ) , Since it consist highest frequency l = 120 h = class size = 20 f 1 = frequency of modal class = 70 f 0 = frequency of class preceding the modal class = 60 f 2 = frequency of class succeeding the modal class = 40 = 120 + 70 - 60 2 × 70 - 60 - 40 × 20 = 120 + 10 40 × 20 = 120 + 10 2 = 125

The following table gives production yield per hectare of wheat of 100 farms of a village Production yield kg / hec 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 Number of farms 4 6 16 20 30 24 Change the distribution to a 'more than type' distribution and draw its ogive.

The cumulative frequency distribution table by more than type is as below: Production yield kg / hec Number of farms f i Production yield kg / hec Cumulative frequency c . f . 40 - 45 4 More than 40 4 + 96 = 100 45 - 50 6 More than 45 6 + 90 = 96 50 - 55 16 More than 50 16 + 74 = 90 55 - 60 20 More than 55 20 + 54 = 74 60 - 65 30 More than 60 30 + 24 = 54 65 - 70 24 More than 65 24 Mark lower class limits on x - axis and cumulative frequency on y - axis. We plot the points 40 , 100 , 45 , 96 , 50 , 90 , 55 , 74 , 60 , 54 , and 65 , 24 . Joining the above points by line segments, the cumulative frequency polygon is as shown below:

The median of the following data is 525 . Find the values of x and y , if total frequency is 100 . Class 0 - 100 100 - 200 200 - 300 300 - 400 400 - 500 500 - 600 600 - 700 700 - 800 800 - 900 900 - 1000 Frequency 2 5 x 12 17 20 y 9 7 4

Cumulative frequency for the given data; Class interval Frequency f Cumulative frequency CF 0 - 100 2 2 100 - 200 5 7 200 - 300 x x + 7 300 - 400 12 x + 19 400 - 500 17 x + 36 500 - 600 20 x + 56 600 - 700 y x + y + 56 700 - 800 9 x + y + 65 800 - 900 7 x + y + 72 900 - 1000 4 x + y + 76 total Σ f = 76 + x + y = 100 Given that the median is 525 . So median class is 500 - 600 Now median = l + N 2 - C F f × h where l = 500 , h = 100 , N = 100 , C F = 36 + x , f = 20 so, 525 = 500 + 100 2 - ( 36 + x ) 20 × 100 ⇒ 525 - 500 = ( 50 - ( 36 + x ) ) × 5 ⇒ 25 = ( 14 - x ) × 5 ⇒ x = 9 Also, Σ f = 76 + x + y = 100 ⇒ 9 + y = 100 - 76 ⇒ y = 15 Hence, x = 9 and y = 15

If the median of the following frequency distribution is 32 . 5 . Find the values of f 1 and f 2 . Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Frequency f 1 5 9 12 f 2 3 2 40

Find the Cumulative Frequency of the given data. class Frequency f Cumulative Frequency c f 0 - 10 f 1 f 1 10 - 20 5 f 1 + 5 20 - 30 9 f 1 + 14 30 - 40 12 f 1 + 26 40 - 50 f 2 f 1 + f 2 + 26 50 - 60 3 f 1 + f 2 + 29 60 - 70 2 f 1 + f 2 + 31 N = ∑ f = 40 Sum of Cumulative Frequency is f 1 + f 2 + 31 = 40 ⇒ f 1 + f 2 = 9 . . . . . . . ( 1 ) Now, The median 32 . 5 is lies between 30 - 40 . Hence, median class is 30 - 40 . ∴ l = 30 , N 2 = 40 2 = 20 , h = 10 , f = 12 , cumulative frequency of preceding class ( cf ) = 14 + f 1 . Now, median is given as, M = l + ( N 2 - cf ) f × h 32 . 5 = 30 + ( 20 - ( 14 + f 1 ) 12 × 10 2 . 5 = ( 60 - 10 f 1 ) 12 30 = 60 - 10 f 1 f 1 = 30 10 f 1 = 3 From equation ( 1 ) , we get f 2 = 9 - f 1 f 2 = 9 - 3 f 2 = 6 Hence, the value of f 1 is 3 and f 2 is 6 .

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The mean of the following frequency distribution is $25$ . Find the value of $f$ .

Class Frequency

$0 - 10$ $5$

$10 - 20$ $18$

$20 - 30$ $15$

$30 - 40$ $f$

$40 - 50$ $6$

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Important Questions on Statistics

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10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>CBSE-2022 (Delhi - II)>Q 19

Find the mean of the following data using the assumed mean method:

Class	Frequency
$0 - 5$	$8$
$5 - 10$	$7$
$10 - 15$	$10$
$15 - 20$	$13$
$20 - 25$	$12$

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>CBSE-2023 (Delhi - I)>Q 21

The empirical relations between the mode, median and mean of a distribution is :

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10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>CBSE-2023 (Delhi - I)>Q 22

The monthly expenditure on milk in $200$ families of Housing Society is given below:

The monthly expenditure in ₹	$1000 - 1500$	$1500 - 2000$	$2000 - 2500$	$2500 - 3000$	$3000 - 3500$	$3500 - 4000$	$4000 - 4500$	$4500 - 5000$
The number of families	$24$	$40$	$33$	$x$	$30$	$22$	$16$	$7$

Find the value of $x$ and also, find the median and mean expenditure on milk.

HARD

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>CBSE-2023 (Delhi - I)>Q 23

The monthly expenditure on milk in $200$ families of a housing society is given below:

Monthly Expenditure(in $₹$ )	$1000 - 1500$	$1500 - 2000$	$2000 - 2500$	$2500 - 3000$	$3000 - 3500$	$3500 - 4000$	$4000 - 4500$	$4500 - 5000$
Number of families	$24$	$40$	$33$	$x$	$30$	$22$	$16$	$7$

Find the value of $x,$ and also find the median and mean expenditure on milk.

EASY

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>CBSE-2023 (Delhi - II)>Q 24

For the following distribution:

Class	$0 - 5$	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$
Frequency	$10$	$15$	$12$	$20$	$9$

The sum of lower limits of median class and model class is:

EASY

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>CBSE-2023 (Delhi - III)>Q 25

For the following distribution:

Marks below	$10$	$20$	$30$	$40$	$50$	$60$
Number of students	$3$	$12$	$27$	$57$	$75$	$80$

The modal class is:

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10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>CBSE-2023 (Outside Delhi)>Q 26

$250$ apples of a box were weighed and the distribution of masses of the apples is given in the following table:

Mass (in grams)	$80 - 100$	$100 - 120$	$120 - 140$	$140 - 160$	$160 - 180$
Number of apples	$20$	$60$	$70$	$x$	$60$

Find the value of $x$ , the mean mass of the apples and the modal mass of the apples.

MEDIUM

10th CBSE

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 14 - Statistics>CBSE-2023 (Outside Delhi)>Q 27

$250$ apples of a box were weighed and the distribution of masses of the apples is given in the following table.

Mass (in $g$ )	$80 - 100$	$100 - 120$	$120 - 140$	$140 - 160$	$160 - 180$
Number of apples	$20$	$60$	$70$	$x$	$60$

Find the modal mass of the apples.

Class	Frequency
$0 - 10$	$5$
$10 - 20$	$18$
$20 - 30$	$15$
$30 - 40$	$f$
$40 - 50$	$6$

The mean of the following frequency distribution is 25. Find the value of f. Class Frequency 0-10 5 10-20 18 20-30 15 30-40 f 40-50 6

Important Questions on Statistics

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The mean of the following frequency distribution is $25$ . Find the value of $f$ .

Class Frequency

$0 - 10$ $5$

$10 - 20$ $18$

$20 - 30$ $15$

$30 - 40$ $f$

$40 - 50$ $6$