The following table gives the ages of 130 members of a club. Age (in years) 20 and under 25 25 and under 30 30 and under 35 35 and under 40 40 and under 45 No. of members 2 ? 29 43 ? If the median age is 37 . 33 years, find the missing frequencies.

We construct the following table as under: Age (in years) 20-25 25-30 30-35 35-40 40-45 No. of member f i 2 f 1 29 43 f 2 Cumulative frequency 2 2 + f 1 31 + f 1 74 + f 1 74 + f 1 + f 2 Since total frequency n = ∑ f i = 130 ⇒ 74 + f 1 + f 2 = 130 ⇒ f 1 + f 2 = 56 . . . . . 1 ∴ n 2 i . e . 65 th observation lies in the class 35 - 40 . Thus, 35 - 40 be the median class. Thus l = 35 ; f = 43 ; c = 31 + f 1 ; i = 5 We know that, Median = l + n 2 - c f × i = 35 + 65 - 31 - f 1 43 × 5 Also given median age = 37 . 33 years 37 . 33 = 35 + 34 - f 1 43 × 5 ⇒ 2 . 33 = 34 - f 1 5 43 ⇒ 100 . 19 = 170 - 5 f 1 ⇒ 5 f 1 = 69 . 81 ⇒ f 1 = 13 . 96 ∴ f 1 = 14 ∴ From 1 ; f 2 = 56 - 14 = 42 Hence, the value of f 1 and f 2 are 14 and 42

Find median, lower quartile and upper quartile for the data set: Variate 15 18 20 22 25 27 30 Frequency 4 6 8 9 7 8 6

We construct the following table: Variate 15 18 20 22 25 27 30 Frequency 4 6 8 9 7 8 6 Cumulative frequency 4 10 18 27 34 42 48 Here , n = ∑ f i = 48 Median = n 2 th observation + n 2 + 1 th observation 2 = 24 th value + 25 th value 2 = 22 + 22 2 = 22 [Since both 24 th and 25 th values are same] Q 1 = Lower quartile = n + 1 4 th value = 48 + 1 4 th value i.e 12 . 25 th value = 12 th value + 0 . 25 13 th value - 12 th value = 20 + 0 . 25 20 - 20 = 20 ∴ Q 1 = 20 Upper quartile = Q 3 = 3 n + 1 4 th value = 3 48 + 1 4 th value = 36 . 75 th value = 36 th value + 0 . 75 37 th value - 36 th value = 27 + 0 . 75 27 - 27 = 27 Hence, the median, lower quartile and upper quartile are 22 , 20 and 27

Find median, lower quartile and upper quartile for the data set: Variate 25 31 34 40 45 48 50 60 Frequency 3 8 10 15 10 9 9 2

We construct the following table: Variate 25 31 34 40 45 48 50 60 Frequency 3 8 10 15 10 9 6 2 Cumulative frequency 3 11 21 36 46 55 61 63 Here, n = ∑ f i = 63 (odd) Median = n + 1 2 th observations = 63 + 1 2 th observations i.e. 32 th observations = 40 Q 1 = Lower quartile = n + 1 4 th value = 63 + 1 4 th value i.e 16 th value = 34 ∴ Q 1 = 34 Upper quartile = Q 3 = 3 n + 1 4 th value = 3 63 + 1 4 th value = 48 th value = 48 Hence, the median, lower quartile and upper quartile are 40 , 34 and 48 .

E M B I B E

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>EXERCISE>Q 17

HARD

11th ICSE

IMPORTANT

Earn 100

Find the values of $a$ and $b$ from the following data:

Variable 10-20 20-30 30-40 40-50 50-60 60-70

Frequency 4 28 a 20 b 16

Given that median is $41.5$ and total frequency is $100$ .

Important Questions on Statistics

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>EXERCISE>Q 18

The following table gives the ages of $130$ members of a club.

Age (in years)	20 and under 25	25 and under 30	30 and under 35	35 and under 40	40 and under 45
No. of members	2	?	29	43	?

If the median age is $37.33$ years, find the missing frequencies.

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>EXERCISE>Q 19

The average monthly pocket money of $10$ children is $₹ 38$ and median $₹ 40$ . What is the total amount of pocket money?

HARD

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>EXERCISE>Q 19

The average monthly pocket money of $10$ children is $₹ 38$ and median $₹ 40$ . If the child getting highest pocket money starts receiving an additional $₹ 20$ per month, what would be the new mean? The new median ?

MEDIUM

11th ICSE

IMPORTANT

Understanding ISC Mathematics Class 11 Volume 2>Chapter 1 - Statistics>EXERCISE>Q 1

Find median, lower quartile and upper quartile for the data set:

$9, 5, 7, 11, 13, 17, 15$