The median of the following data is 525 . Find x and y if the sum of all frequencies is 100 . Class 200 - 300 300 - 400 400 - 500 500 - 600 600 - 700 700 - 800 Frequency 16 x 17 20 15 y

We observe that the classes of marks are continuous and are already arranged in ascending order. First, we shall find the cumulative frequency (less than type) of the data. Class 200 - 300 300 - 400 400 - 500 500 - 600 600 - 700 700 - 800 Frequency 16 x 17 20 15 y Frequency (cumulative) 16 16 + x 33 + x 53 + x 68 + x 68 + x + y Given that the sum of all frequencies is 100 . Hence, 68 + x + y = 100 ⇒ x + y = 32 ----- (i) Also, given that the median of the following data is 525 . This means the median class is 500 - 600 . We know that the median is the value which lies inside the median class and is given by the formula median = l + n 2 - c . f . f × h , where, l = lower limit of the median class n = total number of observations c . f . = cumulative frequency of the class preceding the median class f = frequency of the median class h = size of the class (assuming that all classes have equal size) Here, l = 500 , n = 100 , c . f . = 33 + x , f = 20 and h = 100 . Hence, we have, 500 + 100 2 - 33 + x 20 × 100 = 525 ⇒ 100 2 - ( 33 + x ) 20 × 100 = 25 ⇒ 5 17 - x = 25 ⇒ 17 - x = 5 ⇒ x = 12 From (i), we obtain, x + y = 32 ⇒ 12 + y = 32 ⇒ y = 20 Therefore, the required values are x = 12 and y = 20 .

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

The cumulative frequency table is as below: Class interval Frequency f i Cumulative Frequency c . f . 0 - 10 x x 10 - 20 5 x + 5 20 - 30 9 9 + x + 5 = x + 14 30 - 40 12 12 + x + 14 = x + 26 40 - 50 y x + y + 26 50 - 60 3 3 + x + y + 26 = x + y + 29 60 - 70 2 2 + x + y + 29 = x + y + 31 N = ∑ f i = x + y + 31 Given that, ∑ f i = 40 ⇒ x + y + 31 = 40 ⇒ x + y = 40 - 31 = 9 . . . . . . . ( 1 ) We have, N = 40 So, N 2 = 40 2 = 20 Since the median is 32 . 5 which lies in the class interval 30 - 40 . Thus, the median class is 30 - 40 . Here, l = 30 (the lower limit of the median class) f = 12 (frequency of the median class) h = 10 (width of each class) N = 40 (number of observations) and C = x + 14 (cumulative frequency of class preceding the median class) ∴ Median = l + N 2 - C f × h ⇒ 32 . 5 = 30 + 20 - x + 14 12 × 10 ⇒ 32 . 5 - 30 = 20 - x - 14 6 × 5 ⇒ 2 . 5 × 6 = 6 - x × 5 ⇒ 15 = 30 - 5 x ⇒ 5 x = 30 - 15 ⇒ x = 15 5 = 3 From ( 1 ) , we get x + y = 9 ⇒ y = 9 - x = 9 - 3 = 6 Hence, the value of x = 3 and y = 6 .

The following table gives the daily income of 50 workers of a factory: Daily income (in ₹ ) 100 - 120 120 - 140 140 - 160 160 - 180 180 - 200 Number of workers 12 14 8 6 10 Find the mean, median and mode of the above data.

We have the following: Daily income Mid-value ( x i ) Frequency ( f i ) Cumulative frequency ( f i × x i ) 100 - 120 110 12 12 1320 120 - 140 130 14 26 1820 140 - 160 150 8 34 1200 160 – 180 170 6 40 1020 180 - 200 190 10 50 1900 Ʃ f i = 50 Ʃ f i × x i = 7260 Mean, x ¯ = ∑ i f i × x i ∑ i f i = 7260 50 = 145 . 2 Now, N = 50 ⇒ N 2 = 25 The cumulative frequency just greater than 25 is 26 and the corresponding class is 120 - 140 . Thus, the median class is 120 - 140 . ∴ l = 120 , h = 20 , f = 14 , c = c f of preceding class = 12 and N 2 = 25 Now, we have, median, M e = l + h × N 2 - c f = 120 + 20 × 25 - 12 14 = 120 + 20 × 13 14 = 138 . 57 Now, we know that mode = 3 (median) - 2 (mean) = 3 × 138 . 57 – 2 × 145 . 2 = 125 . 31 Therefore, the required mean, median and mode are 145 . 2 , 138 . 57 and 125 . 31 , respectively.

100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows: Number of letters 1 - 4 4 - 7 7 - 10 10 - 13 13 - 16 16 - 19 Number of surnames 6 30 40 16 4 4 Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

The frequency distribution table is given below: Number of letters Midpoint x i Number of letters (frequency) f i f i x i Cumulative frequency c f 1 - 4 2 . 5 6 2 . 5 × 6 = 15 6 4 - 7 5 . 5 30 5 . 5 × 30 = 165 6 + 30 = 36 7 - 10 8 . 5 40 8 . 5 × 40 = 340 36 + 40 = 76 10 - 13 11 . 5 16 11 . 5 × 16 = 184 76 + 16 = 92 13 - 16 14 . 5 4 14 . 5 × 4 = 58 92 + 4 = 96 16 - 19 17 . 5 4 17 . 5 × 4 = 70 96 + 4 = 100 TOTAL Σ f i = 100 Σ f i x i = 832 N = 100 For median: We have, total frequency, N = 100 ⇒ N 2 = 100 2 = 50 Observe, c f = 76 is just greater than 50 Thus, median class = 7 - 10 Median is given by, median = L + N 2 - c f f × h where, L = Lower class limit of median class = 7 N 2 = 50 c f = Cumulative frequency of the class preceding median class = 36 f = Frequency of the median class = 40 h = Class interval of the median class = 3 Substituting these values in the formula of median, we get, Median = 7 + 50 - 36 40 × 3 ⇒ Median = 7 + 14 40 × 3 ⇒ Median = 7 + 1 . 05 ⇒ Median = 8 . 05 For mean: Mean is given by, Mean = ∑ f i x i ∑ f i ⇒ Mean = 832 100 [Refer the values from the table given above] ⇒ Mean = 8 . 32 For mode: Here, the highest frequency is 40 So, the modal class = 7 - 10 Mode is given by, mode = L + f 1 - f 0 2 f 1 - f 0 - f 2 × h where, L = Lower class limit of the modal class = 7 h = Class interval of the modal class = 3 f 1 = Frequency of the modal class = 40 f 0 = Frequency of the class preceding the modal class = 30 f 2 = Frequency of the class succeeding the modal class = 16 Substituting values in the formula of mode, we obtain, mode = 7 + 40 - 30 2 × 40 - 30 - 16 × 3 ⇒ Mode = 7 + 10 34 × 3 ⇒ Mode = 7 + 0 . 88 ⇒ Mode = 7 . 88 Thus, median is 8 . 05 , mean is 8 . 32 and mode is 7 . 88

Find the mean, mode and median (up to two decimal places) for the following data: Class 5 - 15 15 - 25 25 - 35 35 - 45 45 - 55 55 - 65 65 - 75 Frequency 2 3 5 7 4 2 2

We have the following: Class Mid-value ( x i ) Frequency ( f i ) Cumulative frequency ( f i × x i ) 5 - 15 10 2 2 20 15 - 25 20 3 5 60 25 - 35 30 5 10 150 35 - 45 40 7 17 280 45 - 55 50 4 21 200 55 - 65 60 2 23 120 65 - 75 70 2 25 140 Ʃ f i = 25 Ʃ f i × x i = 970 Mean, x ¯ = ∑ i f i × x i ∑ i f i = 970 25 = 38 . 8 Now, N = 25 ⇒ N 2 = 12 . 5 The cumulative frequency just greater than 25 is 17 and the corresponding class is 35 - 45 . Thus, the median class is 35 - 45 . ∴ l = 35 , h = 10 , f = 7 , c = c . f . of preceding class = 10 and N 2 = 12 . 5 Now, we have, median, M = l + h × N 2 - c f = 35 + 10 × 12 . 5 - 10 7 = 35 + 10 × 2 . 5 7 = 35 + 3 . 57142 . . . . . . . . = 35 + 3 . 57 [Up to two places of decimal places] = 38 . 57 Now, we know that mode = 3 (median) - 2 (mean) = 3 × 38 . 57 – 2 × 38 . 8 = 115 . 71 - 77 . 6 = 38 . 11 Therefore, the required mean, median and mode are 38 . 8 , 38 . 57 and 38 . 11 , respectively.

M L Aggarwal Solutions for Chapter: Statistics, Exercise 4: Exercise 15.4

Author:M L Aggarwal

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Statistics, Exercise 4: Exercise 15.4

Attempt the practice questions on Chapter 15: Statistics, Exercise 4: Exercise 15.4 with hints and solutions to strengthen your understanding. CBSE Syllabus Standard Mathematics for Class X solutions are prepared by Experienced Embibe Experts.

Questions from M L Aggarwal Solutions for Chapter: Statistics, Exercise 4: Exercise 15.4 with Hints & Solutions

EASY

10th CBSE

IMPORTANT

CBSE Syllabus Standard Mathematics for Class X>Chapter 15 - Statistics>Exercise 15.4>Q 3

Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.

HARD

10th CBSE

IMPORTANT

CBSE Syllabus Standard Mathematics for Class X>Chapter 15 - Statistics>Exercise 15.4>Q 12

A life insurance agent found the following data for the distribution of ages of $100$ policyholders. Calculate the median age, if policies are only given to persons having age $18$ years onwards but less than $60$ years. If the median age is $k$ years, find $k$ up to two places of decimal.

Age in years	Number of policy holders
Below $20$	$2$
Below $25$	$6$
Below $30$	$24$
Below $35$	$45$
Below $40$	$78$
Below $45$	$89$
Below $50$	$92$
Below $55$	$98$
Below $60$	$100$

HARD

10th CBSE

IMPORTANT

CBSE Syllabus Standard Mathematics for Class X>Chapter 15 - Statistics>Exercise 15.4>Q 13

Find the median (up to one place of decimal) for the following frequency distribution:

Class interval	$10 - 19$	$20 - 29$	$30 - 39$	$40 - 49$	$50 - 59$	$60 - 69$	$70 - 79$
Frequency	$2$	$4$	$8$	$9$	$4$	$2$	$1$

HARD

10th CBSE

IMPORTANT

CBSE Syllabus Standard Mathematics for Class X>Chapter 15 - Statistics>Exercise 15.4>Q 14

The median of the following data is $525$ . Find $x$ and $y$ if the sum of all frequencies is $100$ .

Class	$200 - 300$	$300 - 400$	$400 - 500$	$500 - 600$	$600 - 700$	$700 - 800$
Frequency	$16$	$x$	$17$	$20$	$15$	$y$

HARD

10th CBSE

IMPORTANT

CBSE Syllabus Standard Mathematics for Class X>Chapter 15 - Statistics>Exercise 15.4>Q 15

If the median of the following frequency distribution is $32.5$ . Find the values of $x$ and $y$ .

Class	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	Total
Frequency	$x$	$5$	$9$	$12$	$y$	$3$	$2$	$40$

HARD

10th CBSE

IMPORTANT

CBSE Syllabus Standard Mathematics for Class X>Chapter 15 - Statistics>Exercise 15.4>Q 16

The following table gives the daily income of $50$ workers of a factory:

Daily income (in $₹$ )	$100 - 120$	$120 - 140$	$140 - 160$	$160 - 180$	$180 - 200$
Number of workers	$12$	$14$	$8$	$6$	$10$

Find the mean, median and mode of the above data.

HARD

10th CBSE

IMPORTANT

CBSE Syllabus Standard Mathematics for Class X>Chapter 15 - Statistics>Exercise 15.4>Q 17

$100$ surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Number of letters	$1 - 4$	$4 - 7$	$7 - 10$	$10 - 13$	$13 - 16$	$16 - 19$
Number of surnames	$6$	$30$	$40$	$16$	$4$	$4$

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames.

HARD

10th CBSE

IMPORTANT

CBSE Syllabus Standard Mathematics for Class X>Chapter 15 - Statistics>Exercise 15.4>Q 18

Find the mean, mode and median (up to two decimal places) for the following data:

Class	$5 - 15$	$15 - 25$	$25 - 35$	$35 - 45$	$45 - 55$	$55 - 65$	$65 - 75$
Frequency	$2$	$3$	$5$	$7$	$4$	$2$	$2$

M L Aggarwal Solutions for Chapter: Statistics, Exercise 4: Exercise 15.4

M L Aggarwal Mathematics Solutions for Exercise - M L Aggarwal Solutions for Chapter: Statistics, Exercise 4: Exercise 15.4

Questions from M L Aggarwal Solutions for Chapter: Statistics, Exercise 4: Exercise 15.4 with Hints & Solutions

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