Shown below are number line representation of the medians of two different data sets. Which of the two representations of median could be correct?

Cannot say without knowing the elements of the data sets

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.

If the median of 60 observations, given below is 28 . 5 , find the values of x and y . Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Frequency 5 x 20 15 y 5

Class interval frequency ( f i ) Cumulative frequency ( c f ) 0 - 10 5 5 10 - 20 x 5 + x 20 - 30 20 25 + x 30 - 40 15 40 + x 40 - 50 y 40 + x + y 50 - 60 5 45 + x + y Total ∑ f i = 60 or 45 + x + y Given that the total number of observations is 60 and the value of the median is 28 . 5 . We have to find the value of x and y . So, we have, 5 + x + 20 + 15 + y + 5 = 60 ⇒ x + y + 45 = 60 ⇒ y = 60 - 45 - x ⇒ y = 15 - x -----(i) 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). Since the median is 28 . 5 , the median class is 20 - 30 Here, n = 60 , l = 20 , c f = 5 + x , f = 20 and h = 30 - 20 = 10 So, the median = l + n 2 - c f f × h = 20 + 60 2 - 5 + x 20 × 10 = 20 + 30 - 5 - x 20 × 10 = 20 + 25 - x 2 . So, 20 + 25 - x 2 = 28 . 5 ⇒ 25 - x 2 = 28 . 5 - 20 ⇒ 25 - x 2 = 8 . 5 ⇒ 25 - x = 8 . 5 × 2 = 17 ⇒ - x = 17 - 25 ⇒ - x = - 8 ⇒ x = 8 From (i), we get, y = 15 - x = 15 - 8 = 7 Therefore, the values of x and y are 8 and 7 , respectively.

EASY

10th CBSE

IMPORTANT

Earn 100

A ketchup manufacturing company sells ketchup in three different bottle sizes. Now, it wants to sell only one size in the market. It has data on how the three sizes perform in the market.

Based on which measure of central tendency, should the company fix the size of the ketchup bottle? Justify your answer.

Important Questions on Statistics

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10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 7

A survey was conducted on $80$ gamers on how many games did they play in a day. The data is given below.

Number of games	Number of gamers
$1 - 2$	$20$
$2 - 3$	$24$
$3 - 4$	$10$
$4 - 5$	$12$
$5 - 6$	$8$
$6 - 7$	$4$
$7 - 8$	$2$

Which of the following is the modal class?

EASY

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 8

Shown below are number line representation of the medians of two different data sets.

Question Image

Which of the two representations of median could be correct?

MEDIUM

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 9

In statistics, an outlier is a data point that differs significantly from other observations of data set. If an outlier is included in the following data set, which measure(s) of central tendency would change?

$12, 15, 22, 44, 44, 48, 50, 51$

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 10

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

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 11

$100$ 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.

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 12

If the median of $60$ observations, given below is $28.5$ , find the values of $x$ and $y$ .

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$
Frequency	$5$	$x$	$20$	$15$	$y$	$5$

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 13

Find the missing frequencies in the following frequency distribution table, if the total frequency is $100$ and mode is $46 \frac{2}{3}$ .

Class	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$
Frequency	$5$	$8$	$7$	$x$	$28$	$20$	$10$	$y$

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 14

A TV reporter was given a task to prepare a report on the rainfall of the city Dispur of India in a particular year. After collecting the data, he analysed the data and prepared a report on the rainfall of the city. Using this report, he drew the following graph for a particular time period of $66$ days.

Question Image

Obtain the mode of the data if mean rainfall is $23.4 c m$

Important Questions on Statistics

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