Samacheer Kalvi 10th Maths Solutions Chapter 8 Unit Exercise 8 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 (Unit Exercise 8) |
Chapter Name | Statistics and Probability |
TNSCERT Class 10th Maths Pdf | all Unit Exercise Solution
Unit Exercise – 8
(1) The mean of the following frequency distribution is 62.8 and the sum of all frequencies is 50. Compute the missing frequencies f1 and f2.
Class Interval | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
Frequency | 5 | f1 | 10 | f2 | 7 | 8 |
Solution:
Given, that N = 50 and mean (Σfixi/N) = 62.8
Class interval | Middle value (xi) | Frequency (fi) | Fixi |
0 – 20 | 10 | 5 | 50 |
20 – 40 | 30 | F1 | 30f1 |
40 – 60 | 50 | 10 | 500 |
60 – 80 | 70 | F2 | 70f2 |
80 – 100 | 90 | 7 | 630 |
100 – 120 | 110 | 8 | 88 |
Σfi = 30 + f1+f2 | Σfixi = 2060+20f1+70f2 |
Now, N = Σfi
50 = 30 + f1 + f2
f1 + f2 = 20 —– (i)
Now, mean = Σfixi/N
62.8 = 2060+30f1+70f2/50
2060 + 30f1 + 70f2 = 3140
30f1 + 70f2 = 1080
3f1 + 7f2 = 108 —- (ii)
Now, (i) × 3 (ii) × 1, we get
3f1 + 3f2 = 60
3f1 + 7f2 = 108
(-) (-) (-)
________________
-4f2 = -48
f2 = 48/4
f2 = 12
From (i), Putting f2 = 12, we get,
f1 + f2 = 20
f1 + 12 = 20
f1 = 20 – 12
f1 = 8
∴ f1 = 8 and f2 = 12
Therefore the missing frequency f1 and f2 are value is 8 and 12.
(2) The diameter of circles (in mm) drawn in a design are given below.
Diameters | 33-36 | 37-40 | 41-44 | 45-48 | 49-52 |
Number of circles | 15 | 17 | 21 | 22 | 25 |
Calculate the standard deviation.
Solution:
Let mean (A) = 42.5
Diameters | Middle value (xi) | Number of circles (fi) | di = xi-A
di = xi-42.5 |
fidi | Fidi2 |
33 – 36 | 34.5 | 15 | -8 | -120 | 960 |
37 – 40 | 38.5 | 17 | -4 | -68 | 272 |
41 – 44 | 42.5 | 21 | 0 | 0 | 0 |
45 – 48 | 46.5 | 22 | 4 | 88 | 352 |
49 – 52 | 50.5 | 25 | 8 | 200 | 1600 |
Σfi = N = 100 | Σfidi = 100 | Σfidi2 = 3184 |
Now N = 100
∴ Standard deviation (σ) = √Σfidi2/N – (Σfidi/N)2
σ = √3184/100 – (100/100)2
σ = √31.84 – 1
σ = √30.84
σ = 5.55
Therefore, the standard deviation (σ) is 5.55.
(3) The frequency distribution is given below.
X | K | 2k | 3k | 4k | 5k | 6k |
f | 2 | 1 | 1 | 1 | 1 | 1 |
In the table, k is a positive integer, has a varience of 160. Determine the value of k.
Solution:-
Let, the mean (A) = 3k and c = k
xi | fi | di = xi-A
di = xi – 3K |
di = xi-3k/k | fidi | fidi2 |
K | 2 | -2k | -2 | -4 | 8 |
2k | 1 | -k | -1 | -1 | 1 |
3k | 1 | 0 | 0 | 0 | 0 |
4k | 1 | K | 1 | 1 | 1 |
5k | 1 | 2k | 2 | 2 | 4 |
6k | 1 | 3k | 3 | 3 | 9 |
2fi = N
= 7 |
Σfidi = 1 | Σfidi2 = 23 |
Now, N = 7 and c = k
Standard deviation (σ) = c ×√Σfidi2/N – (Σfidi/N)2
∴Varience (σ2) = c2× (Σfidi2/N – (Σfidi/N)2)
Now, 160 = k2× {23/7 – (1/7)2}
160 = k2× (161-1/49)
160 = k2× 160/49
k2 = 160×49/160
k2 = 49
k = 7
Therefore, the value of k = 7
(4) The standard deviation of some temperature data in degree celsius (°C) is 5. If the data were converted into degree Farenheit (°F) then what is the variance?
Solution:
Given that, standard deviation of some temperature data in degree Celsius (°c) = 5
We know that,
C/5 = F-32/9
Now, c/5 = F/9 – 32/9
C/5 = F/9
[∵we know that standard deviation will not change any fixed constant add or subtract]
Now, standard deviation (c) (σ) = 5
∴ c = 5 then
5/5 = F/9
F = 9
∴ Standard deviation (°F) = 9
∴ Fahrenheit degree this temperature data variance = (9)2 = 81.
Thus, the required variance is 81.
(5) If for a distribution,Σ(x-5) = 3, Σ(x – 5)2 = 43, and total number of observations is 18, find the mean and standard deviation.
Solution:
Given that, Σ(x-5) = 3 and N = 18
Σ(x – 5)2 = 43
∴We know that mean = Σx/N
Now, Σ(x-5) = 3
Σx – Σ5 = 3
Σx – 5×18 = 3
Σx – 90 = 3
Σx = 93 —- (i)
∴mean (x) = Σx/N = 93/18 = 5.166
= 5.17
Standard deviation (σ) = √Σ(x)/N – (Σ(x)/N)2
Now, Σ(x – 5)2 = 43
Σ (x2 – 10x + 25) = 43
Σ x2 – Σ10x + Σ25 = 43
Σx2 – 10Σx + 25 Σ1 = 43
Σx2 – 10 × 93 + 25 × 18 = 43 [By equation (i) and N = 18]
Σx2 – 930 + 450 = 43
Σx2 = 43 + 930 – 450
Σx2 = 523
Now, standard deviation
σ = √Σx2/N – (Σx/N)2
σ = √523/18 – (5.17)2 [Σx/N = x = 5.17]
σ = √24.05 – 26.72
σ = √2.33
σ = 1.526
σ = 1.53
Therefore, the mean (x) = 5.17 and standard deviation (σ) = 1.53
(6) Prices of peanut packets in various places of two cities are given below. In which city, prices were more stable?
Prices in city A | 20 | 22 | 19 | 23 | 16 |
Prices in city B | 10 | 20 | 18 | 12 | 15 |
Solution:
We find to two cities coefficient of variation of prices. Then compare which city, prices were more stable.
Now, find the coefficient variation of prices in city A.
Let mean (A) = 20
xi | di = xi – A
= xi – 20 |
di2 |
20 | 0 | 0 |
22 | 2 | 4 |
19 | -1 | 1 |
23 | 3 | 9 |
16 | -4 | 16 |
Σxi = 100 | Σdi = 0 | Σdi2 = 30 |
∴ n = 5 then mean (x) = Σxi/n = 100/5
= 20
Standard deviation (σ1) = √Σdi2/n – (Σdi/n)
σ1 = √30/5 – (0)2
σ1 = √6
σ1 = 2.45
∴ Now, coefficients of variation in city A (c.v1) = σ1/x × 100%
= 2.45/20 × 100%
= 245/20 %
= 49/4 %
= 12.25%
Them, find the coefficients of variation in city B.
Let mean (A) = 15
xi | di = xi–A
= xi – 15 |
di2 |
10 | -5 | 25 |
20 | 5 | 25 |
18 | 3 | 9 |
12 | -3 | 9 |
15 | 0 | 0 |
Σxi = 75 | Σdi = 0 | Σdi2 = 68 |
Now, N = 5 then mean (x) = Σxi/n
= 75/5
= 15
Standard deviation (σ2) = √Σdi2/n – (Σdi/n)2
σ2= √68/5 – (0)2
σ2 = √13.6
σ2 = 3.69
Coefficients of variation in city B
(C. v2) = σ2/x × 100%
= 3.69 /16 × 100%
= 369/15 %
= 24.6 %
Now, 12.25% ∠24%
i.e (C.v1) ∠(c.v2)
Thus, city A, prices more stable.
(7) If the range and coefficient of range of the data are 20 and 0.2 respectively, then find the largest and smallest values of the data.
Solution:
Let, the largest value = L
Let, the smallest value = s
Now, given that range (R) = 20
∴ L – S = 20 —- (i)
and coefficient of range = 0.2
L-s/L+S = 0.2
L + S = 20/0.2 [By equation (i)]
L + S = 100 — (ii)
Now, (i) + (ii), we get
L – S + L + S = 20 + 100
2L = 120
L = 60
From (i), L = 60 putting, we get
L – S = 20
60 – S = 20
S = 60 – 20
S = 40
Thus, the value of largest and smallest is 60 and 40 respectively.
(8) If two dice are rolled, then find the probability of getting the product of face value 6 or the difference of face values 5.
Solution:
Sample space = {(1, 1), (1, 2), (1, 3), (1, 4), (1 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
n (s) = 36
Let, A1 be the event of getting the product of face value 6.
A1 = {(2, 3), (1, 6), (3, 2), (6, 1)}
n (A1) = 4
∴ P (A1) = n (A1)/n(s) = 4/36 = 1/9
Let, A2 be the event of getting the difference of face values 5.
A2 = {(1, 6), (6, 1)}
n (A2) = 2
∴ P (A2) = n(A2)/n(s) = 2/36 = 1/18
Now, A1∩A2 = {(1, 6), (6, 1)}
∴n (A1∩A2) = 2
∴ P (A1∩A2) = n(A1∩A1)/n(S) = 2/36 = 1/18
Now, P (A1∪A2)
= P (A1) + P (A2) – P (A1∩A2)
= 1/9 + 1/18 – 1/18
= 1/9
Therefore, the probability of getting the product of face value B or the difference of face values 5 is 1/9.
(9) In a two children family, find the probability that there is at least one girl in a family.
Solution:
Sample space = {BB, BG, GB, GG}
n (S) = 4
Let, A be the event of getting at least one girl in the family.
A = {BG, GB, GG}
n (A) = 3
P (A) = n(A)/n(S) = 3/4
Thus, the probability of least one girl in this family is 3/4.
(10) A bag contains 5 white and some black balls. If the probability of drawing a black ball from the bag is twice the probability of drawing a white ball then find the number of black balls.
Solution:
Let, the number of black ball = x
Sample space = 5+x
n (s) = 5+x
Let, A be the event of drawing a black ball.
n (A1) = x
f (A1) = n(A)/n(S) = x/5+x
Let, A2 be the event of drawing a white ball.
∴n (A2) = 9
P (A2) = n(A2)/n(s) = 5/5+x
Now, P (A1) = 2 P(A2)
x/5+x = 2× 5/5+x
x/5+x = 10/5+x
x = 10
Therefore, the number of black balls in the bag is 10.
(11) The probability that a student will pass the final examination in both English and Tamil is 0.5 and the probability of passing neither is 0.1. If the probability of passing the English examination is 0.75, what is the probability of passing the Tamil examination?
Solution:
Let, A and B be the event of getting students English and Tamil Pass.
Given that, P (A∩B) = 0.5, P (A) = 0.75
P (A∪B) = 0.1
1 – P (A∪B) = 0.1
P (A∪B) = 1-0.1 = 0.9
∴ P (A∪B) = 0.9
Now, P (A∪B) = 0.9
P (A) + P(B) – P (A∩B) = 0.9
0.75 + P(B) – 0.50 = 0.9
P (B) + 0.25 = 0.90
P(B) = 0.90 – 0.25
P(B) = 0.65
P(B) = 65/100
P(B) = 13/20
Therefore, the probability of passing the tamil examination is 13/20.
(12) The King, Queen and Jack of the suit spade are removed from a deck of 52 cards. One card is selected from the remaining cards. Find the probability of getting
(i) a diamond
(ii) a queen
(iii) a spade
(iv) a heart card bearing the number 5.
Solution:
Total card no = 52
Now, king, Queen and jack of the suit spade card are removed.
∴n (S) = 52-3
n (s) = 49
(i) Let, A be the event of getting a diamond
∴n (A1) = 13
P (A1) = n(A1)/n(s) = 13/49
∴ P (a diamond) = 13/49
(ii) Let, A2 be the event of getting a queen now, ohe queen removed.
So, n (A2) = 3
∴ P (A2) = n(A1)/n(s) = 3/49
P (a queen) = 3/49
(iii) Let, A3 be the event of getting a spade.
∴Now, king, queen and jack spade card removed
So, n(A3) = 13 – 3 = 10
∴ P (A3) = n(A3)/n(s) = 10/49
P (a spade) = 10/49
(iv) Let, A4 be the event of getting a heart card bearing the number 5.
∴n (A4) = 1
∴ P (A4) = n(A4)/n(s) = 1/49
P (a heart card bearing the number 5) = 1/49
Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 8 Statistics and Probability Unit Exercise 8
To get more exercise solutions for TN Samacheer Kalvi Class 10 Math, click below
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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