AP Board Class 10 Maths Chapter 14 Statistics Ex 14.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 14 Statistics Ex 14.3 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

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AP Board Class 10 Maths Chapter 14 Statistics Ex 14.3 Textbook Solutions PDF: Download Andhra Pradesh Board STD 10th Maths Chapter 14 Statistics Ex 14.3 Book Answers

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Andhra Pradesh Board Class 10th Maths Chapter 14 Statistics Ex 14.3 Textbooks Solutions PDF

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Andhra Pradesh State Board Class 10th Maths Chapter 14 Statistics Ex 14.3 Books Solutions

Board	AP Board
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	10th
Subject	Maths
Chapters	Maths Chapter 14 Statistics Ex 14.3
Provider	Hsslive

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10th Class Maths 14th Lesson Statistics Ex 14.3 Textbook Questions and Answers

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:

Sum of the frequencies = 68
∴ 𝑛2 = 682 = 34
Hence median class=125 – 145
Lower boundary of the median class, l = 125
cf – cumulative frequency of the class preceding the median class=22
f – frequency of the median class=20
h = class size = 20
Median = 𝑙+[n2−cf]f×h
= 125 + [34−22]20 × 20
∴ Median = 125 + 12 = 137
Maximum number of consumers lie in the class 125 – 145
Modal class is 125 -145
l – lower limit of the modal class=125
f₁ – frequency of the modal class=20
f₀ – frequency of the class preceding the modal class=13
f₂ – frequency of the class succeeding the modal class=14
h – size of the class=20
Mode (Z) = 𝑙+𝑓1−𝑓0(𝑓1−𝑓0)+(𝑓1−𝑓2)×ℎ
Mode (Z) = 125 + 20−13(20−13)+(20−14)×20
= 125 + 77+6 × 20
= 125 + 14013
= 125 + 10.76923
∴ Mode = 135.769
Mean x⎯⎯⎯=a+ΣfiuiΣfi×h
a = assumed mean = 135
∴ 𝐱⎯⎯⎯ = 135 + 768
= 135 + 0.102941
≃ 135.1
Mean, Median and Mode are approximately same in this case.

Question 2.
If the median of 60 observations, given below is 28.5, find the values of x and y.

Answer:

Median = 𝑙+[n2−cf]f×h
It is given that ∑f = n = 60
So, 45 + x + y = 60
x + y = 60 – 45 = 15
x + y = 15 ….. (1)
The median is 28.5 which lies be-tween 20 and 30.
Median class=20 – 30
Lower boundary of the median class ‘l’ = 20
𝑁2 = 602 = 30
cf – cumulative frequency = 5 + x
h = 10
Median = 𝑙+[n2−cf]f×h
⇒ 28.5 = 20 + 30−5−𝑥20 × 10
⇒ 28.5 = 20 + 25−𝑥2
25−𝑥2 = 28.5 – 20 = 8.5
25 – x = 2 × 8.5
x = 25- 17 = 8
also from (1); x + y = 15
8 + y = 15
y = 7
∴ x = 8; y = 7.

Question 3.
A life insurance agent found the following data about distribution of ages of 100 policy holders. Calculate the median age. [Policies are given only to persons having age 18 years onwards but less than 60 years.

Answer:

The given distribution being of the less than type, 25, 30, 35, give the upper limits of corresponding class intervals. So the classes should be 20 – 25, 25 – 30, 30 – 35, ………. 55 – 60.
Observe that from the given distribution 2 persons with age less than 20.
i.e., frequency of the class below 20 is 2.
Now there are 6 persons with age less than 25 and 2 persons with age less than 20.
∴ The number of persons with age in the interval 20 – 25 is 6 – 2 = 4.
Similarly, the frequencies can be calculated as shown in table.
Number of observations = 100
n = 100
𝑛2 = 1002 = 50, which lies in the class 35-40
∴ 35 – 40 is the median class and lower boundary l = 35
cf = 45;
h = 5;
f = 33
Median = 𝑙+[n2−cf]f×h
= 35 + 50−4533 × 5
= 35 + 533 × 5
= 35 + 0.7575
= 35.7575
∴ Median ≃ 35.76

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:
Since the formula, Median = 𝑙+[n2−cf]f×h assumes continuous classes assumes continuous class, the data needs to be converted to continuous classes.
The classes then changes to 117.5 – 126.5; 126.5 – 133.5, …… 171.5 – 180.5.

∑f_i = n = 40
𝑛2 = 402 = 20
𝑛2th observation lie in the class 144.5- 153.5
∴ Median class=144.5 – 153.5
Lower boundary, l = 144.5
Frequency of the median class, f = 12
c.f. = 17
h = 9
∴ Median = 𝑙+[n2−cf]f×h
= 144.5 + 20−1712 × 9
= 144.5 + 312 × 9
= 144.5 + 94
= 144.5 + 2.25
∴ Median length = 146.75 mm.

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:

Total observations are n = 400
𝑛2th observation i.e 4002 = 200
200 lies in the class 3000 – 3500
∴ Median class=3000 – 3500
Lower boundary l = 3000
frequency of the median class f = 86
c.f = 130
Class size, h = 500
Median = 𝑙+[n2−cf]f×h
= 3000 + 200−13086 × 500
= 3000 + 7086 × 500
= 3000 + 406.977
= 3406.98
∴ Median life ≃ 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 alphabet 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:
Number of letters in the surnames.
Also find the modal size of the surnames.

Total observations are n = 100
𝑛2 = 1002 = 50
50 lies in the class 7 – 10
∴ Median class=7 – 10
l – lower boundary = 7
f – frequency of the median class=40
cf = 36
Class size h = 3
Median:
Median = 𝑙+[n2−cf]f×h
= 7 + 50−3640 × 3
= 7 + 1440 × 3
= 7 + 4240
= 7 + 1.05
= 8.05
∴ Median = 8.05.

Mean:
Assumed mean, a = 8.5
Mean x⎯⎯⎯=a+ΣfidiΣfi
= 8.5 + (−18)100
= 8.5 – 0.18
= 8.32
∴ Mean = 8.32.

Mode:
Maximum number of surnames = 40
∴ Modal class=7-10
l – lower boundary of the modal class=7
Mode (Z) = 𝑙+𝑓1−𝑓0(𝑓1−𝑓0)+(𝑓1−𝑓2)×ℎ
l = 7; f₁ = 40, f₀ = 30, f₂ = 16, h = 3
Mode (Z) = 7 + 40−30(40−30)+(40−16) × 3
= 7 + 1010+24 × 3
= 7 + 3034
= 7 + 0.882
= 7.882

Median = 8.0.5; Mean = 8.32; Modal size = 7.88.

Question 7.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.

Answer:

Number of observations (n) = ∑f_i
𝑛2 = 302 = 15
15 lies in the class 50 – 55
∴ Median class=50-55
l – lower boundary of the median class=55
f – frequency of the median class=8
c.f = 5
Class size h = 6
Median = 𝑙+[n2−cf]f×h
= 50 + 15−58 × 6
= 50 + 7.5
= 57.5
= 50 + 7.5 = 57.5
∴ Median weight = 57.5 kg.

AP Board Textbook Solutions PDF for Class 10th Maths

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Andhra Pradesh Board Class 10th Maths Chapter 14 Statistics Ex 14.3 Textbooks for Exam Preparations

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Thursday, June 9, 2022