# AP 10th Class Maths 13th Chapter Statistics Exercise 13.2 Solutions

Well-designed AP 10th Class Maths Guide Chapter 13 Statistics Exercise 13.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Statistics Class 10 Exercise 13.2 Solutions – 10th Class Maths 13.2 Exercise Solutions

Question 1.
The following table shows the ages of the patients admitted in a hospital during a year.

 Age (in years) 5-15 15-25 25-35 35-45 45-55 55 – 65 No. of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution:
Mode :

Note : Highest frequency class can be treated as the modal class.
$$\text { Mode }=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] h$$
l = lower limit of modal class = 35
f1 = frequency of the modal class = 23
f0 = frequency of the class preceding the modal class = 21
f2 = frequency of the class succeeding the modal class = 14
h = height of the class = 10

$$\text { Mode }=35+\left(\frac{23-21}{2 \times 23-21-14}\right) \times 10$$
= 35 + $$\frac{2}{46-35}$$ × 10 = 35 + $$\frac{20}{11}$$ = 35 + 1.81
∴ Mode = 36.81

Mean:

Note : Highest frequency class can be treated as the mean class
Mean=$$\frac{\Sigma \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\Sigma \mathrm{f}_{\mathrm{i}}}$$
a = class mark of the assumed mean class = 40
xi = class mark
h = height of the class = 10
Σfi = Total frequency = 80
Σfiui = -37
Mean $$\overline{\mathrm{x}}$$ = 40 + $$\frac{-37}{80}$$ × 10 = 40 – $$\frac{37}{8}$$
= 40 – 4.625
Mean = 35.375
∴ Mode is more than mean.

Question 2.
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:

Determaine the model of lifetimes of components.
Solution:

Note : Highest frequency class can be treated as the modal class.
$$\text { Mode }=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] h$$
l = lower limit of modal class = 60
f1 = frequency of the modal class = 61
f0 = frequency of the class preceding the modal class = 52
f2 = frequency of the class succeeding the modal class = 38
h = height of the class = 20
$$\text { Mode }=60+\left[\frac{61-52}{2 ×62-52-38}\right]$$ × 20
= 60 + ($$\frac{90×20}{32}$$)
= 60 + 5.625
= 65.625
∴ Modal lifetime of the components = 65.625 hours

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 ₹) Number of families 1000-1500 24 1500-2000 40 2000-2500 33 2500-3000 28 3000-3500 30 3500-4000 22 4000-4500 16 4500-5000 7

Solution:

Note : Highest frequency class can be treated as the modal class
$$\text { Mode }=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] h$$
l = lower limit of modal class = 1500
f1 = frequency of the modal class = 40
f0 = frequency of the class preceding the modal class = 24
f2 = frequency of the class succeeding the modal class = 33
$$\text { Mode }=1500+\left[\frac{40-24}{2×40-24-33}\right]$$ × 500
= 1500 + $$\frac{16×500}{23}$$
= 1500 + 347.82 = 1847.82 /-
∴ Modal monthly expenditure = ₹ 1847.82 /-

Note : Highest frequency class can be treated as the mean class

Mean $$\overline{\mathbf{x}}=a+\frac{\sum f_i u_i}{\sum f_i} \times h$$
a = class mark of the assumed mean class = 1750
xi = class mark
h = height of the class = 500
Σfi = Total frequency = 200
Σfiui = 365

= 1750 + 912.50
Mean = ₹ 2662.50
Therefore, mean monthly expenditure = ₹ 2662.50

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

Solution:
Mean :

Note : Highest frequency class can be treated as the mean class

Mean $$\overline{\mathbf{x}}=a+\frac{\sum f_i u_i}{\sum f_i} \times h$$
a = class mark of the assumed mean class = 32.5
xi = class mark
h = height of the class = 5
Σfi = Total frequency = 35
Σfiui = -23

Therefore, mean number of students per each teacher = 30.625 (or) 31 (Approximately)

Mode:

Note : Highest frequency class can be treated as the mean class
$$\text { Mode }=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] h$$
l = lower limit of the modal class = 30
f1 = frequency of the modal class = 10
f0 = frequency of the class preceding the modal class = 9
f2 = frequency of the class succeeding the modal class = 3
h = height of the class = 5

$$\text { Mode }=30+\left[\frac{10-9}{2×10-9-3}\right]$$ × 5
= 30 + $$\frac{1}{20-12}$$ × 5
= 30 + $$\frac{1}{8}$$ = 30 + 0.625
Mode = 30.625
Therefore, modal number of students per each teacher = 30.625 (or) 31 (Approximately)

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 scored Number of batsmen 3000-4000 4 4000-5000 18 5000-6000 9 6000-7000 7 7000-8000 6 8000-9000 3 9000-10000 1 10000-11000 1

Find the mode of the data.
Solution:

Note : Highest frequency class can be treated as the mean class
$$\text { Mode }=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] \times h$$
l = lower limit of the modal class = 4000
f1 = frequency of the modal class = 18
f0 = frequency of the class preceding the modal class = 4
f2 = frequency of the class succeeding the modal class = 9
h = height of the class = 1000
$$\text { Mode }=4000+\left[\frac{18-4}{2×18-4-9}\right]$$ × 1000
= 4000 + $$\frac{14}{36-13}$$ × 1000 = 4000 + ($$\frac{14}{23}$$) × 1000
= 4000 + 608.69 = 4608.69
Mode = 4608.69 (or) 4609 runs (approx.)
Therfore, mode = 4609 runs approximately.

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:

Solution:

Note : Highest frequency class can be treated as the modal class.
$$\text { Mode }=l+\left[\frac{f_1-f_0}{2 f_1-f_0-f_2}\right] \times h$$
l = lower limit of the modal class = 40
f1 = frequency of the modal class = 20
f0 = frequency of the class preceding the modal class = 12
f2 = frequency of the class succeeding the modal class = 11
h = height of the class = 10
$$\text { Mode }=40+\left[\frac{20-12}{2×20-12-11}\right]$$ × 10
= 40 + $$\frac{8}{40-23}$$ × 10 = 40 + $$\frac{80}{17}$$ = 40 + 4.706 = 44.706 cars
Therefore, mode of the data = 44.706 cars.