Question 4 - Case Based Questions (MCQ) - Chapter 13 Class 10 Statistics

Last updated at April 16, 2024 by Teachoo

Question

The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching center are given as follows.

Speed (in Km/h)

85 − 100

100 − 115

115 − 130

130 − 145

No. of Players

11

9

8

5

Question 1

What is the modal class of the given data?

(a) 85-100

(b) 100-115

(c) 115-130

(d) 130-145

Question 2

What is the value of class interval for the given data set?

(a) 10

(b) 15

(c) 5

(d) 20

Question 3

What is the median class of the given data?

(a) 85-100 (b) 100-115

(c) 115-130 (d) 130-145

Question 4

What is the median of bowling speed?

(a) 109.17 km/hr (Approx)

(b) 109.71 km/hr (Approx)

(c) 107.17 km/hr (Approx)

(d) 109.19 km/hr (Approx)

Question 5

What is the sum of lower limit of modal class and upper limit of median class?

(a) 100

(b) 200

(c) 300

(d) 400

Transcript

Question The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching center are given as follows. Question 1 What is the modal class of the given data? (a) 85-100 (b) 100-115 (c) 115-130 (d) 130-145 Modal class is class with highest frequency
Since 85 – 100 has highest frequency 11
∴ It is the modal class
So, the correct answer is (a)
Question 2 What is the value of class interval for the given data set? (a) 10 (b) 15 (c) 5 (d) 20 Class interval
= Upper class limit – Lower class limit
= 100 – 85
= 15
So, the correct answer is (b)
Question 3 What is the median class of the given data? (a) 85-100 (b) 100-115 (c) 115-130 (d) 130-145 Here,
𝑵/𝟐=33/2= 16.5
∴ 100 – 115 is the median class
So, the correct answer is (a)
Question 4 What is the median of bowling speed? (a) 109.17 km/hr (Approx) (b) 109.71 km/hr (Approx) (c) 107.17 km/hr (Approx) (d) 109.19 km/hr (Approx)Median = l + (𝑵/𝟐 −𝒄𝒇)/𝒇 × h
∴ 100 – 115 is the median class
And,
l = lower limit of median class = 100
h = class-interval = 15
cf = cumulative frequency of the class before median class
= 11
f = frequency of the median class
= 9
Putting values in formula
Median = l + (𝑁/2 −𝑐𝑓)/𝑓 × h
= 100 + (𝟏𝟔.𝟓 −𝟏𝟏)/𝟗 × 15
= 100 + 5.5/9 × 15
= 100 + 27.5/3
= 100 + 9.17
= 109.17 km/hr (approx.)
So, the correct answer is (a)
Question 5 What is the sum of lower limit of modal class and upper limit of median class? (a) 100 (b) 200 (c) 300 (d) 400 Modal Class is 85 – 100
Lower limit of Modal Class = 85
Median Class is 100 – 115
Upper limit of Median Class = 115
Sum of lower limit of modal class and upper limit of median class
= 85 + 115 = 200
So, the correct answer is (B)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

Hi, it looks like you're using AdBlock :(

Displaying ads are our only source of revenue. To help Teachoo create more content, and view the ad-free version of Teachooo... please purchase Teachoo Black subscription.

Please login to view more pages. It's free :)

Teachoo gives you a better experience when you're logged in. Please login :)

Solve all your doubts with Teachoo Black!

Teachoo answers all your questions if you are a Black user!