Students can Download Class 10 Maths Chapter 13 Statistics Ex 13.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka State Syllabus Class 10 Maths Chapter 13 Statistics Ex 13.3

Question 1.

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

Answer:

Median

Cumulative frequency (cf) > \(\frac{N}{2}\) = 34 is 42 class interval corresponding to 42 is 125 – 145

∴ Median class is 125 – 145.

l = 125, h = 20, f = 20, cf = 22

Median = 137

Mean:

Assumed mean = a = 135, h = 20

= 135 + 2.06

= 137.06

Mode:

Modal class is 125 – 145. It has maximum number of frequencies.

l = 125, h = 20, f_{1} = 20, f_{0} = 13 and f_{2} = 14

∴ On comparison, all the three measures are approximately equal in this case.

Use this median calculator to quickly find the median of a set of numbers. Just enter your numbers in the box and click on the button that says calculate.

Question 2.

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

Answer:

Σf_{i} = N = 60 (Given)

45 + x + y = 60

x + y = 60 – 45 = 15

x + y = 15 → (1)

The median is 28.5, which lies in the class 20 – 30

∴ Median class 20 – 30

l = 20, h = 10, f = 20, cf = 5 + x

25 – x = 17

x = 25 – 17

x = 8

Put x = 8 in equation (1),

x + y = 15

y = 15 – x = 15 – 8

y = 7

Question 3.

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

Answer:

Cumulative frequency distribution table

Median class is 35 – 40

l = 35, h = 5, f = 33 and cf = 45

= 35 + 0.76

= 35.76

Thus, the median age = 35.76 years

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:

Find the median length of the leaves.

(Hint: The data needs 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.)

Answer:

The given class interval data is in inclusive method.

∴ Convert to exclusive method.

Subtractive 0.5 for lower limit and add 0.5 for upper limit

∴ Median class is 144.5 – 153.5

l = 144.5, h = 9, f = 12 and cf= 17

= 144.5 + 2.25

= 146.75

∴ Median length of the leaves is 146.75mm.

Question 5.

The following table gives the distribution of the life time of 400 neon lamps:

Find the median life time of a lamp.

Answer:

Cumulative frequency > \(\frac{N}{2}\) = 200 is 216.

Median class is 3000 – 3500

l = 3000, h = 500, f = 86 and cf = 130

= 3000 + 406.98

= 3406.98 hours.

The median life time of a lamp is 3406.98 hours.

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 alphabets in the surnames was obtained as follows:

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.

Answer:

Median:

Cumulative frequency > \(\frac{N}{2}\) = 50 is 76.

Median class internal is 7 – 10

l = 7 , h = 3, f = 40, cf = 36

= 7 + 1.05 = 8.05

The median number of letters in the surnames is 8.05.

Mean:

Assumed mean = 8.5

Class size = h = 3

= 8.5 – 0.18

Mean = 8.32

∴ The mean number of letters in surnames is 8.32

Mode:

Modal class is 7 – 10. It has maximum number of frequency.

l = 7 , h = 3, f_{i} = 40, f_{0} = 30 and f_{2} = 16

= 7 + \(\frac{10}{34}\) × 3 = 7 + \(\frac{30}{34}\)

= 7 + 0.88

= 7.88

∴ The modal size of the sur names is 7.88.

Question 7.

The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:

Cumulative frequency > \(\frac{N}{2}\) = 15 is 19

∴ Median class is 55 – 60

l = 55, h = 5, f = 6 and cf = 13

= 55 + \(\frac {2}{6}\) × 5

= 55 + \(\frac {5}{3}\)

= 55 + 1.67

= 56.67

∴ The median weight of the student is 56.67 kg