Find the median marks in the data. Marks 15 20 10 25 5 Number of students 10 8 6 4 1

First arrange the observations given in ascending. Marks Number of students Cumulative frequency 5 1 1 10 6 7 15 10 17 20 8 25 25 4 29 Total = 29 Here, N = 29 . 29 is odd. Median = N + 1 2 th observation = 29 + 1 2 th = 15 th observation From the table corresponding values of 15 th observations is the 15 . Hence, the median of the data is 15 .

Find the mean, median and mode of the following data: Classes 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 120 - 140 Frequency 6 8 10 12 6 5 3

Here, we have Class Frequency ( f i ) x i f i x i Cumulative frequency 0 - 20 6 10 60 6 20 - 40 8 30 240 14 40 - 60 10 50 500 24 60 - 80 12 70 840 36 80 - 100 6 90 540 42 100 - 120 5 110 550 47 120 - 140 3 130 390 50 N = ∑ f = 50 ∑ f i x i = 3120 Here, total frequency is N = 50 . So, median class is N 2 = 50 2 = 25 . The cumulative frequency just greater than 25 is 36 , then the median class is 60 - 80 , such that the lower limit ( l ) = 60 . Cumulative frequency of the class preceding median class is ( C ) = 24 , frequency of the median class = 12 , class size ( h ) = 20 . So, the median is l + N 2 - C f × h = 60 + 25 - 24 12 × 20 = 60 + 20 12 = 61 . 67 Thus, the median is 61 . 67 . Mean of the data = ∑ f i x i ∑ f Therefore, the mean = 3120 50 = 62 . 4 Now, mode of the = l + f k - f k - 1 2 f k - f k - 1 - f k + 1 × h l = lower limit of modal class h = width of the class interval f k = frequency of the modal class f k - 1 = frequency of the class preceding the modal class f k + 1 = frequency of the class succeeding the modal class Here, l = 60 , f k = 12 , f k - 1 = 10 , f k + 1 = 6 and h = 20 . So, the mode = 60 + 12 - 10 24 - 10 - 6 × 20 = 60 + 2 8 × 20 = 60 + 5 = 65 Hence, the required mode is 65 , median is 61 . 67 and mean is 62 . 4 .

100 surnames were randomly picked up from a local telephone directory. The frequency distribution of the number of letters in the English alphabet 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 up to two places after decimal point.

We have to find the mean, the median and the mode of the given data. For the mean we shall use the step-deviation method. The frequency distribution table is given below. Number of letters Number of surnames ( f i ) Cumulative frequency ( c f ) Classmark ( x i ) d i = x i - a u i = x i - a h f i u i 1 - 4 6 6 2 . 5 - 6 - 2 - 12 4 - 7 30 36 5 . 5 - 3 - 1 - 30 7 - 10 40 76 8 . 5 0 0 0 10 - 13 16 92 11 . 5 3 1 16 13 - 16 4 96 14 . 5 6 2 8 16 - 19 4 100 17 . 5 9 3 12 Total ∑ f i = 100 ∑ f i u i = - 6 We know that the mean in the step-deviation method is given by x ¯ = a + ∑ f i u i ∑ f i × h . Here, a = 8 . 5 , h = 3 , ∑ f i u i = - 6 and ∑ f i = 100 . So, the mean x ¯ = a + ∑ f i u i ∑ f i × h = 8 . 5 + - 6 100 × 3 = 8 . 5 - 0 . 18 = 8 . 32 Hence, the mean number of letters in the surnames is 8 . 32 . Now, we know that the median for a grouped data is given as median = l + n 2 - c f f × h , where, l = lower boundary of median class, n = number of observations, c f = cumulative frequency of class preceding the median class, f = frequency of median class, h = class size (size of the median class). Here, n = 100 . So, n 2 = 100 2 = 50 , which belongs to the class 7 - 10 . Hence, the median class is 7 - 10 So, in this case, n = 100 , l = 7 , c f = 36 , f = 40 and h = 10 - 7 = 3 So, the median = l + n 2 - c f f × h = 7 + 50 - 36 40 × 3 = 7 + 1 . 05 = 8 . 05 Hence, the median number of letters in the surnames is 8 . 05 Now, for grouped data, the mode is calculated using the following formula. Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h , where, l = the lower boundary of the modal class, h = size of the modal class interval, f 1 = frequency of the modal class, f 0 = frequency of the class preceding the modal class, f 2 = frequency of the class succeeding the modal class. Here, the modal class is 7 - 10 [The class with the maximum frequency, called the modal class]. So, here, l = 7 , h = 10 - 7 = 3 , f 1 = 40 , f 0 = 30 and f 2 = 16 . Hence, the mode = 7 + 40 - 30 2 × 40 - 30 - 16 × 3 = 7 + 10 80 - 46 × 3 = 7 + 0 . 88 = 7 . 88 . [up to two places after decimal point] So, the modal size of the surnames is 7 . 88 letters. Therefore, the mean, the median and the mode number of letters in the surnames are 8 . 32 , 8 . 05 and 7 . 88 , respectively.

Embibe Experts Solutions for Chapter: Statistics, Exercise 1: Exercise

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Statistics, Exercise 1: Exercise

Attempt the practice questions on Chapter 14: Statistics, Exercise 1: Exercise with hints and solutions to strengthen your understanding. Mathematics Crash Course (Based on Revised Syllabus-2023) solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Statistics, Exercise 1: Exercise with Hints & Solutions

MEDIUM

10th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Statistics>Exercise>Q 13

Can mode be calculated for grouped data with unequal class sizes?

EASY

10th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Statistics>Exercise>Q 14

Find the mode of the following data.

$20, 3, 7, 13, 3, 4, 6, 7, 19, 15, 7, 18, 3$

EASY

10th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Statistics>Exercise>Q 15

Does the mode change, if another observation is added to the data $0, 1, 2, 2, 2, 3, 3, 4, 5, 6$ . Comment.

HARD

10th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Statistics>Exercise>Q 19

Find the median marks in the data.

Marks	$15$	$20$	$10$	$25$	$5$
Number of students	$10$	$8$	$6$	$4$	$1$

HARD

10th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Statistics>Exercise>Q 20

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

Classes	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$	$100 - 120$	$120 - 140$
Frequency	$6$	$8$	$10$	$12$	$6$	$5$	$3$

HARD

10th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Statistics>Exercise>Q 21

The following frequency distribution gives the monthly consumption of electricity of $68$ consumers of a locality. Find the median, mean, and mode of the data and compare them.

Monthly consumption (in units)	$65 - 85$	$85 - 105$	$105 - 125$	$125 - 145$	$145 - 165$	$165 - 185$	$185 - 205$
Number of consumers	$4$	$5$	$13$	$20$	$14$	$8$	$4$

HARD

10th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Statistics>Exercise>Q 22

$100$ surnames were randomly picked up from a local telephone directory.

Question Image

The frequency distribution of the number of letters in the English alphabet 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 up to two places after decimal point.

EASY

10th Andhra Pradesh Board

IMPORTANT

Mathematics Crash Course (Based on Revised Syllabus-2023)>Chapter 14 - Statistics>Exercise>Q 42

$X -$ coordinate of intersecting point of two ogives, represents

Embibe Experts Solutions for Chapter: Statistics, Exercise 1: Exercise

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Questions from Embibe Experts Solutions for Chapter: Statistics, Exercise 1: Exercise with Hints & Solutions

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