# Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.5 Pdf

## Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.5 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.5: Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 8 Exercise 8.5 – Statistics and Probability on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 8 Exercise 8.5 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 8.5

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

 Board TNSCERT Class 10th Maths Class 10 Subject Maths Chapter 8 (Exercise 8.5) Chapter Name Statistics and Probability

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

Exercise – 8.5

Multiple choice questions

(1) Which of the following is not a measure of dispersion?

(A) Range

(B) Standard deviation

(C) Arithmetic mean

(D) Variance

Ans:

(C) Arithmetic mean

(2) The range of the data 8, 8, 8, 8, 8. . . 8 is

(A) 0

(B) 1

(C) 8

(D) 3

Solution:

Largest value (L) = 8

Smallest value (s) = 8

∴ Range (R) = L – S = 8 – 8 = 0 (A)

0

(3) The sum of all deviations of the data from its mean is

(A) Always positive

(B) Always negative

(C) Zero

(D) Non-zero integer

Solution:

(C) Zero

(4) The mean of 100 observations is 40 and their standard deviation is 3. The sum of squares of all deviations is

(A) 40000

(B) 160900

(C) 160000

(D) 30000

Solution:

Given that,

n = 100, Σxi/n = 40 and σ = 3

σ = √Σx2/n – (Σxi/n)2

σ2 = Σxi2/n – (Σxi/n)2 [Both side square]

9 = Σxi2/100 – (40)2

Σxi2/100 = 1600 + 9 = 1609

Σxi2 = 160900

∴The sum of squares of all deviation is 160900 (B).

(5) Variance of first 20 natural numbers is

(A) 32.25

(B) 44.25

(C) 33.25

(D) 30

Solution:

We know that first natural numbers —- deviation σ = √n2-1/12

n = 20 then σ = √/12

σ = √400-1/12

σ = √399/12

σ = √33.25

Now, variance (σ2) = 33.25 (C)

(6) The standard deviation of a data is 3. If each value is multiplied by 5 then the new variance is

(A) 3

(B) 15

(C) 5

(D) 225

Solution:

New standard deviation = 3×5 = 15

∴ New variance = (15)2 = 225 (D)

(7) If the standard deviation of x, y, z is p then the standard deviation of 3x + 5, 3y + 5, 3z + 5 is

(A) 3p + 5

(B) 3p

(C) p + 5

(D) 9p + 15

Solution:

Standard deviation will not change when any fixed constant add or subtract to the all data.

Now, standard deviation will change when any fixed constant multiple on divided to the all data then new standard deviation previous standard deviation same constant multiple or divided.

Now, 3x + 5, 3y + 5 and 3z + 5

Standard deviation = 3×P = 3P (B)

(8) If the mean and coefficient of variation of a data are 4 and 87.5% then the standard deviation is

(A) 3.5

(B) 3

(C) 4.5

(D) 2.5

Solution:

Given x = 4 and C.V = 87.5

Let, standard deviation = σ

∴ C.V = xσ/x × 100%

87.5 = σ/4 × 100 σ = 3.5

∴ Standard deviation (σ) = 3.5 (A)

(9) Which of the following is incorrect?

(A) P(A) > 1

(B) 0 ≤ (A)1

(C) P() = 0

(D) P(A) + P(A) = 1

Solution:

(A) P (A) > 1

(10) The probability a red marble selected at random from a jar containing p red, q blue and r green marbles is

(A) q/p+q+r

(B) p/p+q+r

(C) p+q/p+q+r

(D) p+r/p+q+r

Solution:

Total marbles = P + q + r

∴ n (S) = p + q + r

Red marbles = p

n (Red marbles) = p

p (red marble) = n (red marble)/n (s) = p/p+q+r (B)

(11) A page is selected at random from a book. The probability that the digit at units place of the page number chosen is less than 7 is

(A) 3/10

(B) 7/10

(C) 3/9

(D) 7/9

Solution:

Sample space = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

∴ n (s) = 10

Let, A be the event of getting the page number chosen is less than 7.

A = {0, 1, 2, 3, 4, 5, 6}

n (A) = 7

P (A) = n(A)/n(S) = 7/10 (B)

(12) The probability of getting a job for a person is x/3. If the probability of not getting the job is 2/3 then the value of x is

(A) 2

(B) 1

(C) 3

(D) 1.5

Solution:

Let, A be the event of getting a job given that P(A) = x/3 and P(A) = 2/3

Now, P(A) + P(A) = 1

x/3 + 2/3 = 1

x+2/3 = 1

x + 2 = 3

x = 3 – 2 = 1 (B)

(13) Kamalam went to play a lucky draw contest. 135 tickets of the lucky draw were sold. If the probability of Kamalam winning is 1/9, then the number of tickets bought by kamalam is

(A) 5

(B) 10

(C) 15

(D) 20

Solution:

Let, A be the event of getting the number of tickets bought by kamalam.

∴ Given P(A) = 1/9 and n(S) = 135

∴ P(A) = n(A)/n(S)

1/9 = n(A)/135

∴ n(A) = 135/9 = 15 (C)

(14) If a letter is chosen at random from the English alphabets {a, b,…,z}, then the probability that the letter chosen precedes x

(A) 12/13

(B) 1/13

(C) 23/26

(D)  3/26

Solution:

Sample space = {a, b, e, —- z}

n (S) = 26

Let, A be the event of getting the letter chosen precedes x

∴ A = {a, b, c, —- w}

n (A) = 23

∴ P (A) = n (A)/n (S) = 23/26 (C)

(15) A purse contains 10 notes of ₹2000, 15 notes of ₹500, and 25 notes of ₹200. One note is drawn at random. What is the probability that the note is either a ₹500 note or ₹200 note?

(A) 1/5

(B) 3/10

(C) 2/3

(D) 4/5

Solution:

Example space = 10 + 15 + 25 n (s) = 50

Let, A be the event getting a ₹500 note

Let B be the event of getting a ₹200 note

n (A) = 15 n (B) = 25

P (A) = n(A)/n(S) = 15/50

P (B) = n(B)/n(S) = 25/50

Now, n(A∩B) = n (A∩B)/n(s) = 50 = 0

Now, P (A∪B)

= P(A) + P(B) – p (A∩B)

= 15/50 + 25/50 – 0

= 15+25/50

= 40/50

= 4/5 (D)

Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 8 Statistics and Probability Exercise 8.5

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Updated: March 4, 2022 — 2:44 pm