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

The cumulative frequency table is as below: Class interval Frequency f i Cumulative Frequency c . f . 0 - 10 f 1 f 1 10 - 20 5 f 1 + 5 20 - 30 9 9 + f 1 + 5 = f 1 + 14 30 - 40 12 12 + f 1 + 14 = f 1 + 26 40 - 50 f 2 f 1 + f 2 + 26 50 - 60 3 3 + f 1 + f 2 + 26 = f 1 + f 2 + 29 60 - 70 2 2 + f 1 + f 2 + 29 = f 1 + f 2 + 31 N = ∑ f i = f 1 + f 2 + 31 Given that, ∑ f i = 40 ⇒ f 1 + f 2 + 31 = 40 ⇒ f 1 + f 2 = 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 = f 1 + 14 (cumulative frequency of class preceding the median class) ∴ Median = l + N 2 - C f × h ⇒ 32 . 5 = 30 + 20 - f 1 + 14 12 × 10 ⇒ 32 . 5 - 30 = 20 - f 1 - 14 6 × 5 ⇒ 2 . 5 × 6 = 6 - f 1 × 5 ⇒ 15 = 30 - 5 f 1 ⇒ 5 f 1 = 30 - 15 ⇒ f 1 = 15 5 = 3 From ( 1 ) , we get f 1 + f 2 = 9 ⇒ f 2 = 9 - f 1 = 9 - 3 = 6 Hence, the value of f 1 = 3 and f 2 = 6 .

The following table gives the frequency distribution of married women by age at marriage: Age (in years) Frequency 15 - 19 53 20 - 24 140 25 - 29 98 30 - 34 32 35 - 39 12 40 - 44 9 45 - 49 5 50 - 54 3 55 - 59 2 60 and above 2 Calculate the median and interpret the results.

The given frequency distribution is not continuous. We convert it into continuous by subtracting the conversion factor from the lower limit and adding it to the upper limit of each class interval. Here, the conversion factor = 20 - 19 2 = 1 2 = 0 . 5 The cumulative frequency table is as below: Age (in years) Class Interval Frequency f i Cumulative Frequency c . f . 15 - 19 14 . 5 - 19 . 5 53 53 20 - 24 19 . 5 - 24 . 5 140 140 + 53 = 193 25 - 29 24 . 5 - 29 . 5 98 98 + 193 = 291 30 - 34 29 . 5 - 34 . 5 32 32 + 291 = 323 35 - 39 34 . 5 - 39 . 5 12 12 + 323 = 335 40 - 44 39 . 5 - 44 . 5 9 9 + 335 = 344 45 - 49 44 . 5 - 49 . 5 5 5 + 344 = 349 50 - 54 49 . 5 - 54 . 5 3 3 + 349 = 352 55 - 59 54 . 5 - 59 . 5 2 2 + 352 = 354 60 and above 59 . 5 and above 2 2 + 354 = 356 Total N = ∑ f i = 356 We have, N = 356 So, N 2 = 356 2 = 178 The cumulative frequency just greater than 178 is 193 , which corresponds to the class 19 . 5 - 24 . 5 Thus, the median class is 19 . 5 - 24 . 5 . Here, l = 19 . 5 (the lower limit of the median class) f = 140 (frequency of the median class) h = 5 (width of each class) N = 356 (number of observations) and C = 53 (cumulative frequency of class preceding the median class) ∴ Median = l + N 2 - C f × h = 19 . 5 + 178 - 53 140 × 5 = 19 . 5 + 125 140 × 5 = 19 . 5 + 4 . 46 = 23 . 96 approx Hence, the median is 23 . 96 years and nearly half of the women married between the age group 19 . 5 - 24 . 5 .

Find the median for the following data: Marks Number of students Less than 10 0 Less than 30 10 Less than 50 25 Less than 70 43 Less than 90 65 Less than 110 87 Less than 130 96 Less than 150 100

Here, we are given less than cumulative frequency table. First, we prepare the frequency table: Marks Class interval Number of students c . f . Frequency f i Less than 10 0 - 10 0 0 Less than 30 10 - 30 10 10 - 0 = 10 Less than 50 30 - 50 25 25 - 10 = 15 Less than 70 50 - 70 43 43 - 25 = 18 Less than 90 70 - 90 65 65 - 43 = 22 Less than 110 90 - 110 87 87 - 65 = 22 Less than 130 110 - 130 96 96 - 87 = 9 Less than 150 130 - 150 100 100 - 96 = 4 We have, N = 100 So, N 2 = 100 2 = 50 The cumulative frequency just greater than 50 is 65 , which corresponds to the class 70 - 90 Thus, the median class is 70 - 90 . Here, l = 70 (the lower limit of the median class) f = 22 (frequency of the median class) h = 20 (width of each class) N = 100 (number of observations) and C = 43 (cumulative frequency of class preceding the median class) ∴ Median = l + N 2 - C f × h = 70 + 50 - 43 22 × 20 = 70 + 7 11 × 10 = 70 + 6 . 36 = 76 . 36 approx Hence, the median is 76 . 36 .

Form a grouped distribution and calculate the median: Marks Number of students Above 0 80 Above 10 76 Above 20 71 Above 30 65 Above 40 57 Above 50 43 Above 60 28 Above 70 15 Above 80 10 Above 90 8 Above 100 0

Here, we have given more than cumulative frequency table. First, we prepare the frequency table: Marks Class interval Number of students Frequency f i Cumulative frequency c . f . Above 0 0 - 10 80 80 - 76 = 4 4 Above 10 10 - 20 76 76 - 71 = 5 5 + 4 = 9 Above 20 20 - 30 71 71 - 65 = 6 6 + 9 = 15 Above 30 30 - 40 65 65 - 57 = 8 8 + 15 = 23 Above 40 40 - 50 57 57 - 43 = 14 14 + 23 = 37 Above 50 50 - 60 43 43 - 28 = 15 15 + 37 = 52 Above 60 60 - 70 28 28 - 15 = 13 13 + 52 = 65 Above 70 70 - 80 15 15 - 10 = 5 5 + 65 = 70 Above 80 80 - 90 10 10 - 8 = 2 2 + 70 = 72 Above 90 90 - 100 8 8 - 0 = 8 8 + 72 = 80 Above 100 100 - 110 0 0 0 + 80 = 80 We have, N = 80 So, N 2 = 80 2 = 40 The cumulative frequency just greater than 40 is 52 , which corresponds to the class 50 - 60 Thus, the median class is 50 - 60 . Here, l = 50 (the lower limit of the median class) f = 15 (frequency of the median class) h = 10 (width of each class) N = 80 (number of observations) and C = 37 (cumulative frequency of class succeeding the median class) ∴ Median = l + N 2 - C f × h = 50 + 40 - 37 15 × 10 = 50 + 3 3 × 2 = 50 + 2 = 52 Hence, the median is 52 .

Find the mean, mode and median of the following data: Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Total Frequency 8 16 36 34 6 100

We prepare the following frequency distribution table to find mean, mode and median: Class interval Frequency f i Cumulative frequency c . f . Class Mark x i Deviations d i = x i - 25 u i = d i h = x i - a 10 f i u i 0 - 10 8 8 0 + 10 2 = 10 2 = 5 5 - 25 = - 20 - 20 10 = - 2 - 16 10 - 20 16 16 + 8 = 24 15 - 10 - 1 - 16 20 - 30 36 36 + 24 = 60 25 = a 0 0 0 30 - 40 34 34 + 60 = 94 35 10 1 34 40 - 50 6 6 + 94 = 100 45 20 2 12 Total ∑ f i = 100 ∑ f i u i = 14 Let, the assumed mean a = 25 and h = 10 , then we have, Mean = x ¯ = a + Σ f i u i Σ f i × h = 25 + 14 100 × 10 = 25 + 1 . 4 = 26 . 4 Hence, the mean is 26 . 4 . In the given frequency distribution, the class interval 20 - 30 has maximum frequency. So, 20 - 30 is the modal class. Here, l = 20 , f 1 = 36 , f 0 = 16 , f 2 = 34 , h = 10 , Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h = 20 + 36 - 16 2 × 36 - 16 - 34 × 10 = 20 + 20 72 - 50 × 10 = 20 + 20 22 × 10 = 20 + 9 . 09 = 29 . 09 approx Hence, the mode is 29 . 09 . We have, N = 100 so that N 2 = 100 2 = 50 . The cumulative frequency just greater than 50 is 60 , which corresponds to the class 20 - 30 . Thus, the median class is 20 - 30 . Here, l = 20 , f = 36 , h = 10 , N = 100 , and C = 24 ∴ Median = l + N 2 - C f × h = 20 + 50 - 24 36 × 10 = 20 + 26 36 × 10 = 20 + 65 9 = 20 + 7 . 22 = 27 . 22 approx Hence, the median is 27 . 22 .

Find the median of the following frequency distribution Daily wages in 6069 7079 8089 9099 100109 110119 Number of workers 5 15 20 30 20 8

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 9

Calculate the missing frequency from the following distribution, having given that the median of the distribution is $24$ :

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$
Frequency	$5$	$25$	$x$	$18$	$7$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 10

If the median of the following frequency distribution is $32.5$ , find the values of $f_{1}$ and $f_{2}$ :

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	Total
Frequency	$f_{1}$	$5$	$9$	$12$	$f_{2}$	$3$	$2$	$40$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 11

The following table gives the frequency distribution of married women by age at marriage:

Age (in years)	Frequency
$15 - 19$	$53$
$20 - 24$	$140$
$25 - 29$	$98$
$30 - 34$	$32$
$35 - 39$	$12$
$40 - 44$	$9$
$45 - 49$	$5$
$50 - 54$	$3$
$55 - 59$	$2$
$60$ and above	$2$

Calculate the median and interpret the results.

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 12

Find the median for the following data:

Marks	Number of students
Less than $10$	$0$
Less than $30$	$10$
Less than $50$	$25$
Less than $70$	$43$
Less than $90$	$65$
Less than $110$	$87$
Less than $130$	$96$
Less than $150$	$100$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 13

Form a grouped distribution and calculate the median:

Marks	Number of students
Above $0$	$80$
Above $10$	$76$
Above $20$	$71$
Above $30$	$65$
Above $40$	$57$
Above $50$	$43$
Above $60$	$28$
Above $70$	$15$
Above $80$	$10$
Above $90$	$8$
Above $100$	$0$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 14

A life insurance agent found the following data for distribution of ages of $100$ policy holders. Calculate the median age, if policies are given only to persons having age $18$ years onwards but less than $60$ years.

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

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 15

An incomplete distribution is given below:

Variable	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$
Frequency	$12$	$30$	_____	$65$	_____	$25$	$18$

You are given that the median value is $46$ and the total number of items is $230$ .

(i) Using the median formula, fill up the missing frequencies.

(ii) Calculate the A.M. of the completed distribution.

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 16

Find the mean, mode and median of the following data:

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	Total
Frequency	$8$	$16$	$36$	$34$	$6$	$100$

Find the median of the following frequency distribution:

Daily wages (in $₹$ ) $60 - 69$ $70 - 79$ $80 - 89$ $90 - 99$ $100 - 109$ $110 - 119$

Number of workers $5$ $15$ $20$ $30$ $20$ $8$

Important Questions on Statistics

Learn and Prepare for any exam you want !

Daily wages (in $₹$ )	$60 - 69$	$70 - 79$	$80 - 89$	$90 - 99$	$100 - 109$	$110 - 119$
Number of workers	$5$	$15$	$20$	$30$	$20$	$8$

Find the median of the following frequency distribution: Daily wages (in ₹) 60-69 70-79 80-89 90-99 100-109 110-119 Number of workers 5 15 20 30 20 8

Important Questions on Statistics

Learn and Prepare for any exam you want !

Find the median of the following frequency distribution:

Daily wages (in $₹$ ) $60 - 69$ $70 - 79$ $80 - 89$ $90 - 99$ $100 - 109$ $110 - 119$

Number of workers $5$ $15$ $20$ $30$ $20$ $8$