JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.2

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 1
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
The class interval having the maximum frequencies is 35 – 45.
f1 = 23, l = 35, h = 10, f0 = 21, f2 = 14
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 2
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 3
Maximum number of patients admitted in the hospital are of the age 36.8 years. The average age of the patient admitted to the hospital is 35.37 years.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 4
Determine the modal lifetimes of the components.
Class interval having the maximum frequency is 60 – 80.
f1 = 61, f0 = 52, f2 = 38, l = 60, h = 20.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 5

Question 3.
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

Expenditure (in Rs.)No. of families
1000 – 150024
1500 – 200040
2000 – 250033
2500 – 300028
3000 – 350030
3500 – 400022
4000 – 450016
4500 – 50007

Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 6
Step deviation method: \(\bar{x}\) = a + \(\left[\frac{\sum f_i u_i}{\sum f_i}\right]\)h
= 3250 + \(\left[\frac{-235}{200}\right]\) × 500
= 3250 – \(\left[-\frac{1175}{2}\right]\)
= 3250 – 587.5
= 2662.5.
Mean expenditure Rs. 2662.50.
l = 1500, f1 = 40, f2 = 33, f0 = 24, h = 500
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 7
Modal monthly expenditure = 1847.83.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2

Question 4.
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacherNo. of states/U.T.
15 – 203
20 – 258
25 – 309
30 – 3510
35 – 403
40 – 450
45 – 500
50 – 552

Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 8
l = lower limit of the CI = 30, f1 = 10, f0 = 9, f2 = 3, h = 5
Mode = \(l+\left[\frac{\mathrm{f}_1-\mathrm{f}_0}{2 \mathrm{f}_1-\mathrm{f}_0-\mathrm{f}_2}\right] \times \mathrm{h}\)
= \(=30+\left[\frac{10-9}{20-9-3}\right] \times 5\)
= \(30+\left[\frac{1}{8} \times 5\right]=30+\frac{5}{8}\)
= 30 + 0.625 = 30.625.
Most states/UT’s have a student-teacher ratio of 30.6. On an average this ratio is 29.2.

Question 5.
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scoredNumber of batsmen
3000 – 40004
4000 – 500018
5000 – 60009
6000 – 70007
7000 – 80006
8000 – 90003
9000 – 100001
10000 – 110001

Find the mode of the data.
Solution:
l = 4000, f1 = 18, f0 = 4, f2 = 9, h = 1000
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 9

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2

Question 6.
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 10
Solution:
Class interval having the maximum frequency is 40 – 50. f1 = 20, f0 = 12, f2 = 11, l = 40, h = 10.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.2 11

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