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
x_i = class mark
h = height of the class = 10
Σf_i = Total frequency = 80
Σf_iu_i = -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
x_i = class mark
h = height of the class = 500
Σf_i = Total frequency = 200
Σf_iu_i = 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
x_i = class mark
h = height of the class = 5
Σf_i = Total frequency = 35
Σf_iu_i = -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.