# Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.1 Pdf

## Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.1 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.1: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.1 – Statistics and Probability on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 8 Exercise 8.1 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 8.1

### TN Samacheer Kalvi 10th Maths Solutions Chapter 8 – Statistics and Probability

 Board TNSCERT Class 10th Maths Class 10 Subject Maths Chapter 8 (Exercise 8.1) Chapter Name Statistics and Probability

#### TNSCERT Class 10th Maths Pdf | all Exercise Solution

Exercise 8.1

(1) Find the range and coefficient of range of the following data.

(i) 63, 89, 98, 125, 79, 108, 117, 68

(ii) 43.5, 13.6, 18.9, 38.4, 61.4, 29.8

Solution:-

(i) Largest value (l) = 125

Smallest value (s) = 63

∴ Range (R) = L – S = 125 – 63

= 62

Coefficient of range = L-S /L+S

= 62/188

= 0.33

(ii) Largest value (L) = 61.4

Smallest value (S) = 13.6

∴ Range (R) = L – S = 61.4 – 13.6

= 47.8

Coefficient of range = L–S/L+S

= 47.8/75

= 0.64

(2) If the range and the smallest value of a set of data are 36.8 and 13.4 respectively, then find the largest value.

Solution:

Let, the largest value = L

We know that range (R) = L-S

[L = Largest value, S = smallest value]

R = L-S

36.8 = L – 13.4

L = 36.8 + 13.4

L = 50.2

Therefore, the largest value is 50.2.

(3) Calculate the range of the following data.

 Income 400 – 450 450 – 500 500 – 550 550 – 600 600 – 650 Number of workers 8 12 30 21 6

Solution:

Largest value (L) = 650

Smallest value (S) = 400

∴ Range (R) = L-S

= 650-400

= 250

Thus, the rage 250

(4) A teacher asked the students to complete 60 pages of a record note book. Eight students have completed only 32, 35, 37, 30, 33, 36, 35 and 37 pages. Find the standard deviation of the pages completed by them.

Solution:

Given that eight students completes pages are 32, 35, 37, 30, 33, 36, 35 and 37.

Now,

 xi (xi)2 32 1024 35 1225 37 1369 30 900 33 1089 36 1296 35 1225 37 1369 Σxi = 275 Σx2i = 9497

We know that standard deviation

a = √Σx2i/n – (Σxi/n)2

a = √9497/8 – (275/8)2 [∵ n = 8]

a = √1187.125 – 1181.640

a = √5.74

a = 2.34

Therefore, the standard deviation is a = 2.34

(5) Find the variance and standard deviation of the wages of 9 workers given below:

₹310, ₹290, ₹320, ₹280, ₹300, ₹290, ₹320, ₹310, ₹280

Solution:

Now, Let the mean (A) = 300

 xi di = xi – A di = xi – 300 di2 310 10 100 290 -10 100 320 20 400 280 -20 400 300 0 0 290 -10 100 320 20 400 310 10 100 280 -20 400 Σdi = 0 Σdi2 = 2000

Now, n = 9,

We know that variance = Σ di2/n – (Σ di/n)2

σ2 = 2000/9 – (0/9)2

σ2 = 2000/9 = 222.22

Standard deviation σ = √σ 2

= √22.22

= 14.91

Therefore, the variance σ2 = 222.22 and standard deviation σ = 14.91

(6) A wall clock strikes the bell once at 1 o’ clock, 2 times at 2 o’ clock, 3 times at 3 o’ clock and so on. How many times will it strike in a particular day. Find the standard deviation of the number of strikes the bell make a day.

Solution:

A wall clock strikes early morning to afternoon bell are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

Wall clock strikes afternoon to midnight bell are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12

We know, that standard deviation for first n natural numbers

σ = √n2-1/12

Now, n = 12, σ = √(2)2-1/12

σ = √144-1/12

σ = √143/12

σ = √11.92

σ = 3.45

Therefore, the standard deviation of the number of strikes the bell make a day is = 3.45 + 3.45 = 6.90

(7) Find the standard deviation of first 21 natural numbers.

Solution:

We know that standard deviation for first n natural numbers σ = √n2-1/12

Now, n = 21, σ = √(21)2-1/12

= √441-1/12 = √440/12

σ = √36-67

σ = 6.05

Thus, the first 21 natural numbers standard deviation σ = 6.05

(8) If the standard deviation of a data is 4.5 and if each value of the data is decreased by 5, then find the new standard deviation.

Solution:

Given that standard deviation of the data = 4.5 each data decreased by 5.

We know that, the standard deviation will not change when any fixed constant add or subtract to the all values.

Now each data decreased by 5 then the standard deviation of the data = 4.5

(9) If the standard deviation of a data is 3.6 and each value of the data is divided by 3, then find the new variance and new standard deviation.

Solution:

Given that the standard deviation of the data (σ) = 3.

Now, each value of the data is divided by 3. Then the standard deviation (σ) = 36/3 = 1.2

New variance (σ)2 = (1.2)2

= 1.44

New standard deviation (σ) = 1.2

(10) The rainfall recorded in various places of five districts in a week are given below. Find its standard deviation.

 Rainfall (in mm) 45 50 55 60 65 70 Number of places 5 13 4 9 5 4

Solution:

Let mean = 60

 Rain fall (in mm xi) Number of places fi di = xi – A di = xi – 60 di2 fidi fidi2 45 5 -15 225 -75 1125 50 15 -10 100 -130 1300 55 4 -5 25 -20 100 60 9 0 0 0 0 65 5 5 25 25 125 70 4 10 100 40 400 Σfi = 40 Σdi = -15 Σfidi = -160 Σfidi2 = 3050

Now, n = 40,

We know that standard deviation

σ = √Σfidi2/n – (Σfidi/n)2

σ = √3050/40 – (-160/40)2

σ = √76.25 – 16

σ = √60.25 = 7.76

Thus, the standard deviation σ = 7.76

(11) In a study about viral fever, the number of people affected in a town were noted as Find its standard deviation.

 Age in years 0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 50 – 60 60 – 70 Number of people affected 3 5 16 18 12 7 4

Solution:

Let, the mean (A) = 35 and C = 10

 Age  in years Mid value (xi) Number of people affected (fi) di = xi – A = xi – 35 di = xi-A/10 fidi fidi2 0 – 10 5 3 -30 -3 -9 27 10 – 20 15 5 -20 -2 -10 20 20 – 30 25 16 -10 -1 -16 16 30 – 40 35 18 0 0 0 0 40 – 50 45 12 10 1 12 12 50 – 60 55 7 20 2 14 28 60 – 70 65 4 30 3 12 36 N = 65 Σfidi = 3 Σfidi2 = 139

Now, N = 65

We know that, Standard deviation σ = C × √Σfidi2/N – (Σfidi/N)2

σ = 10× √139/65 – (3/65)2

σ = 10 × √2.1385 – 0.0021

σ = 10 × √2.1364

σ = 10 × 1.4616

σ = 14.616

σ = 14.62

Thus, the standard deviation σ = 14.62

(12) The measurements of the diameters (in cms) of the plates prepared in a factory are given below. Find its standard deviation.

 Diameter (cm) 21 – 24 25 – 28 29 – 32 33 – 36 37 – 40 41 – 44 Number of plates 15 18 20 16 8 7

Solution:

Let, the mean (A) = 34.5 and C = 4.

 Diameter (cm) Mid value (xi) Number of plates fi di = xi – A di = xi-345 di = xi-A/C fidi fidi2 21 – 24 22.5 15 -12 -3 -45 135 25 – 28 26.5 18 -8 -2 -36 72 29 – 32 30.5 20 -4 -1 -20 20 33 – 36 34.5 16 0 0 0 0 37 – 40 38.5 8 4 1 8 8 41 – 44 42.5 7 8 2 14 28 Σfi = N = 84 Σfidi = -79 Σfidi2 = 263

Now, N = 84 and C = 4

Standard deviation σ = C × √Σfidi2/N – (Σfidi/N)2

σ = 4 × √(263/84 – (-79/84)2

σ = 4 × √3.1309 – (0.9404)2

σ = 4 × √3.1309 – 0.8843

σ = 4 × √2.2466

σ = 4 × 1.49

σ = 5.96 = 6

Therefore, the standard deviation σ = 6.

(13) The time taken by 50 students to complete a 100 meter race are given below. Find its standard deviation.

 Time taken (sec) 8.5-9.5 9.5-10.5 10.5-11.5 11.5-12.5 12.5-13.5 Number of students 6 8 17 10 9

Solution:

Let mean (A) = 11

 Time taken (sec) Mid value (xi) Number of student (fi) di = xi – A di = xi – 11 fidi fidi2 8.5 – 9.5 9 6 -2 -12 24 9.5 – 10.5 10 8 -1 -8 8 10.5 – 11.5 11 17 0 0 0 11.5 – 12.5 12 10 1 10 10 12.5 – 13.5 13 9 2 18 36 N = 50 Σfidi = 8 Σfidi2 = 78

Now, N = 50,

Standard deviation σ = √Σfidi2/N – (Σfidi/N)2

σ = √78/50 – (8/50)2

σ = √1.56 – (0.16)2

σ = √1.56 – 0.02

σ = √1.54

σ = 1.24

Therefore, the standard deviation σ = 1.24

(14) For a group of 100 candidates the mean and standard deviation of their marks were found to be 60 and 15 respectively. Later on it was found that the scores 45 and 72 were wrongly entered as 40 and 27. Find the correct mean and standard deviation.

Solution:

Given that n = 100

Mean x = 60 and σ = 15

Now, x = Σx/n

60 = Σx/100

Σx = 6000

Wrong total = 6000

Correct total = 6000 – 40 – 27 + 45 + 7

= 6000 + 5 + 45

= 6050

∴ Correct mean (x) = 6050/100 = 60.5

Wrong standard deviation σ = 15

√Σx2/n – (Σx/n)2 = 15

√Σx2/100 – (60)2 = 15

Σx2/100 – 3600 = 225

Σx2 = 3825 × 100

Now, correct value of Σx2 = 382500 – (40)2 – (27)2 + (45)2 + (72)2

= 382500 – 1600 – 729 + 2025 + 5184

= 387380

∴ Correct standard deviation σ = √Σx2/n – (Σx/n)2

σ = √387380/100 – (60.5)2

σ = √3873.8 – 3660.2

σ = √213.5

∴ σ = √213.5

σ = 14.61

Therefore, the correct mean is 60.5 and correct standard deviation σ = 14.61.

(15) The mean and variance of seven observations are 8 and 16 respectively. If five of these are 2, 4, 10, 12 and 14, then find the remaining two observations.

Solution:

Let remaining two observation are y1 and y2

Now, mean (Σxi/n) = 8 (Given)

Σxi = 8×n

2+4+10+12+14+y1+y2 = 8×7

42 + y1 + y2 = 56

y1 + y2 = 14 —– (i)

Now, given variance

σ2 = Σxi2/12 – (Σxi/n)2

16 = Σxi2/n – (8)2

Σxi2/n = 16 + 64 = 80

Σxi2 = 80×7 [∵ n = 7]

Σxi2 = 560

22 + 42 + 102 + 122 + 142 + y12 + y22 = 560

4 + 16 + 100 + 144 + 196 + y12 + y12 = 560

460 + y12 + y22 = 560

y12 + y22 = 560 – 460

y12 + y22 = 100

Now, (y1 + y2)2 – 2y1y2 = 100

(14)2 – 2y1y2 = 100 [By equation (i)]

196 – 2y1y2 = 100

2y1y2 = 96

y1y2 = 48

∴ y1 = 48/y2

From (i) putting y1 = 48/y2, we get

y1 + y2 = 14

48/y2 + y2 = 14

y22 + 14y2 + 48 = 0

y12 – 8y2 – 6y2 + 48 = 0

y2 (y2 – 8) – 6 (y2 – 8) = 0

(y2 – 8) (y2 – 6) = 0

y2 – 8 = 0

y2 = 8

y2 – 6 = 0

y2 = 6

When, y2 = 8 then,

y1 + y2 = 14

y1 = 14 – 8 = 6

When, y2 = 6 then

y1 + y2 = 14

y1 = 14 – 6 = 8

∴ Thus, the remaining two numbers are 8 and 6 or 6 and 8.

Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 8 Statistics and Probability Exercise 8.1

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Updated: March 4, 2022 — 2:39 pm