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.

The pyramid graph below shows the ages of the 548 Members of Parliament in the 17th Lok Sabha. On an average, how much younger is a female MP than a male MP? Round your answer to the nearest whole number and show your work.

We need to find, on an average how younger is a female MP than a male MP. Here we need to find the average age of male MP and female MP. The first step is to create the table with the midpoint or marks and the product of the frequency and midpoint. To calculate the midpoint we find the average between the class interval . Age(in Years) Class Mark x i Number of female MPs f i f i x i Number of male MPs m i m i x i 25 - 35 30 4 120 7 210 35 - 45 40 17 680 42 1680 45 - 55 50 24 1200 114 5700 55 - 65 60 20 1200 143 8580 65 - 75 70 12 840 125 8750 75 - 85 80 3 240 27 2160 85 - 95 90 0 0 10 900 Total ⅀ f i = 80 ⅀ f i x i = 4280 ⅀ m i = 468 ⅀ m i x i = 27980 The average age of female M.P = 4280 80 = 53 . 5 The average age of male M.P = 27980 468 = 59 . 8 Female M.P is younger than male M.P = 6 . 3 ≈ 6 years. Hence, Female M.P is younger than male M.P by 6 years.

EASY

10th CBSE

IMPORTANT

Earn 100

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

Which of the two representations of median could be correct?

50% studentsanswered this correctly

Important Questions on Statistics

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 $23.4 c m$

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 15

The pyramid graph below shows the ages of the 548 Members of Parliament in the 17th Lok Sabha.

Question Image

On an average, how much younger is a female MP than a male MP? Round your answer to the nearest whole number and show your work.

HARD

10th CBSE

IMPORTANT

Mathematics Textbook of Competency Based Questions for Class X>Chapter 14 - Statistics>Exercise>Q 16

The table below gives the percentage distribution of female teachers in the primary schools of rural areas of various states and union territories (U.T.) of India. Find the mean percentage of female teachers by all three methods.

Percentage of female teachers	$15 - 25$	$25 - 35$	$35 - 45$	$45 - 55$	$55 - 65$	$65 - 75$	$75 - 85$
Number of States/U.T.	$6$	$11$	$7$	$4$	$4$	$2$	$1$

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

Important Questions on Statistics

Learn and Prepare for any exam you want !

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

Which of the two representations of median could be correct?