Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.1 Pdf
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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
| Chapter 1 Relations and Functions | Chapter 2 Numbers and Sequences |
|
Chapter 3 Algebra |
Chapter 4 Geometry |
|
Exercise 4.1 Solution
|
|
Chapter 5 Coordinate Geometry |
Chapter 6 Trigonometry |
| Exercise 5.1 Solution
|
Exercise 6.1 Solution |
|
Chapter 7 Mensuration |
Chapter 8 Statistics and Probability |
| Exercise 7.1 Solution | Exercise 8.1 Solution |
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