The distance (in km) of 35 employees of a company from their residences to the company were found as. 3,10,12,15,11,13,7,31,19,10,12,17,16,2,7,9,7,8,5,18,15,3,12,2,9,6,20,25,15,6,12,7.
(A) Construct a frequency distribution table for the grouped data with equal class width.
(B) Find the percentage of employees whose distance from company to home is less than 5 km.
Answers
Answer:- (A) I have done the construction of the frequency distribution table with equal class width 0-5, 5-10, so on.
(B) The percentage of the number of employees whose home is less than 5Km from the company are = 12.5%
There are 32 employees and the distance from the company to their homes are given in Km. They are
3,10,12,15,11,13,7,31,19,10,12,17,16,2,7,9,7,8,5,18,15,3,12,2,9,6,20,25,15,6,12,7
First, we need to arrange the distance in ascending order
2,2,3,3,5,6,6,7,7,7,7,8,9,9,10,10,11,12,12,12,12,13,15,15,15,16,17,18,19,20,25,31
(A)The Frequency distribution table is:-
Class intervals distance in Km Frequency
0-5 2,2,3,3 4
5-10 5,6,6,7,7,7,7,8,9,9 10
10-15 10,10,11,12,12,12,12,13 8
15-20 15,15,15,16,17,18,19 7
20-25 20 1
25-30 25 1
30-35 31 1
Total Frequency = 32
(B) The number of employees whose home is 5Km from the company are = 4
Total number of employees are = 32
The percentage of the employees whose homes are 5Km from the company is = 4/32×100
= 100/8
= 12.5%
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