JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Jharkhand Board JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 Textbook Exercise Questions and Answers.

JAC Board Class 10 Maths Solutions Chapter 14 Statistics Exercise 14.3

Question 1.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the media, mean and mode of the data and compare them.

Monthly consumption (in units)No. of consumers
65 – 854
85 – 1055
105 – 12513
125 – 14520
145 – 16514
165 – 1858
185 – 2054

Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 1
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 2
Mean is 137 units.
Median is 137 units.
Mode is 135.76 units.
The three measures are approximately the same.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Question 2.
If the median of the distribution given below is 28.5, find the values of x and y.

Class intervalFrequency
0 – 105
10 – 20x
20 – 3020
30 – 4015
40 – 50y
50 – 605
Total60

Solution:

Class intervalFrequencyCumulative frequency
0 – 1055
10 – 20x5 + x
20 – 302025 + x
30 – 401540 + x
40 – 50y40 + x + y
50 – 60545 + x + y
60

n = 60, 45 + x + y = 60
x + y = 60 – 45
x + y = 15
The median is 28.5. It lies in the class interval 20 – 30.
∴ l = 20, f = 20, cf = 5 + x, h = 10
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 3

Question 3.
A life insurance agent found the following data for distribution of ages of 100 policyholders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 years.

Age (in years)No. of policyholders
Below 202
Below 256
Below 3024
Below 3545
Below 4078
Below 4589
Below 5092
Below 5598
Below 60100

Solution:

Class intervalNo. of policyholdersc.f.
Below 2022
20 – 2546
25 – 301824
30 – 352145
35 – 403378
40 – 451189
45 – 50392
50 – 55698
55 – 602100
n = 100\(\frac{n}{2}\) = 50

l = 35, \(\frac{n}{2}\) = 50, cf = 45, f = 33, h = 5
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 4

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Question 4.
The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table:

Length (in mm)Number of leaves
118 – 1263
127 – 1355
136 – 1449
145 – 15312
154 – 1625
163 – 1714
172 – 1802

Find the median length of the leaves.
(Hint: The data need to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 – 126.5, 126.5 – 135.5, ….., 171.5 – 180.5).
Solution:
The data have to be converted to continuous classes for finding the median since the formula. assumes continuous classes.

Class intervalNo. of leavesCumulative frequency (cf)
117.5 – 126.533
126.5 – 135.558
135.5 – 144.5917
144.5 – 153.51229
153.5 – 162.5534
162.5 – 171.5438
171.5 – 180.5240

n = 40, \(\frac{n}{2}\) = 20
The median lies in the class interval 144.5 – 153.5.
l = 144.5, \(\frac{\mathrm{n}}{2}=\frac{40}{2}\) = 20, cf = 17, f = 12, h = 9.
Median = l + \(\left[\frac{\frac{n}{2}-c f}{f}\right]\) × h
= 144.5 + \(\left[\frac{20-17}{12} \times 9\right]\)
= 144.5 + \(\left[\frac{27}{12}\right]\)
= 144.5 + 2.25
= 146.75 mm.

Question 5.
The following table gives the distribution of the lifetime of 400 neon lamps:

Lifetime (in hours)Number of lamps
1500 – 200014
2000 – 250056
2500 – 300060
3000 – 350086
3500 – 400074
4000 – 450062
4500 – 500048

Find the median lifetime of a lamp.
Solution:

Lifetime in hours (CI)No. of lamps (l)Cumulative frequency
1500 – 20001414
2000 – 25005670
2500 – 300060130
3000 – 350086216
3500 – 400074290
4000 – 450062352
4500 – 500048400

The median lies in the class interval 3000 – 3500.
\(\frac{\mathrm{n}}{2}=\frac{400}{2}=200\)
l = 3000, \(\frac{n}{2}\) = 200, cf = 130, f = 86, h = 500.
Median = l + \(\left[\frac{\frac{\mathrm{n}}{2}-\mathrm{cf}}{\mathrm{f}}\right]\) × h
= 3000 + \(\left[\frac{200-130}{86}\right]\) × 500
= 3000 + \(\left[\frac{70}{86}\right]\) × 500
= 3000 + \(\frac{35000}{86}\)
= 3000 + 406.976
Median life of a lamp is 3406.98 hours.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Question 6.
100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabet in the surnames was obtained as follows:

No. of lettersNo. of surnames
1 – 46
4 – 730
7 – 1040
10 – 1316
13 – 164
16 – 194

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.
Solution:
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 5
Hence the modal size of the surnames is 7.88.

JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
JAC Class 10 Maths Solutions Chapter 14 Statistics Ex 14.3 6
Solution:

Weight in kg.No. of studentsCumulative frequency (c.f.)
40 – 4522
45 – 5035
50 – 55813
55 – 60619
60 – 65625
65 – 70328
70 – 75230

\(\frac{n}{2}\) = 15
The median lies in the class 55 – 60.
l = 55, \(\frac{n}{2}\) = 15, c.f. = 13, f = 6, h = 5.
Median = l + \(\frac{1}{2}\) × h
= 55 + \(\left[\frac{15-13}{6}\right]\) × 5
= 55 + \(\frac{2}{6}\) × 5
= 55 + \(\frac{5}{3}\)
= 55 + 1.666 = 56.666
∴ Median = 56.67 kg.
Hence, the median weight of the students is 56.67 kg.

Leave a Comment