Q: Question 3: The following is the probability distribution of X: Find the probability that (i) X is positive (ii) X is non-negative (iii) X is odd (iv) X is even.

(i) P(X is positive) = P(X = 1) + P(X = 2) + P(X = 3) = 0.25 + 0.15 + 0.1 = 0.50

Q: Question 4: The p.m.f. of a r.v. X is given by P(X = x) = x = , for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. Then show that P(X ≤ 2) = P(X ≥ 3).

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = = ………{ }^{n} C_{r}={ }^{n} C_{n-r} = P(X = 5) + P(X = 4) + P(X = 3) = P(X ≥ 3) ∴ P(X ≤ 2) = P(X ≥ 3).

Q: Question 5: In the p.m.f. of r.v. X Find a and obtain c.d.f. of X.

For p.m.f. of a r.v. X ∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1 Let F(x) be the c.d.f. of X. Then F(x) = P(X ≤ x) ∴ F(1) = P(X ≤ 1) = P(X = 1) = F(2) = P(X ≤ 2) = P(X = 1) + P (X = 2) P(3) = P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3) F(4) = P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3)

Q: Question 6: A fair coin is tossed 4 times. Let X denote the number of heads obtained. Write down the probability distribution of X. Also, find the formula for p.m.f. of X.

When a fair coin is tossed 4 times then the sample space is S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT} ∴ n(S) = 16 X denotes the number of heads. ∴ X can take the value 0, 1, 2, 3, 4 When X = 0, then X = {TTTT} ∴ n (X) = 1 ∴ P(X = 0) = When X

Q: Question 7: Find the probability distribution of the number of successes in two tosses of a die, where success is defined as (i) number greater than 4 (ii) six appear on at least one die.

When a die is tossed two times, we obtain (6 × 6) = 36 number of observations. Let X be the random variable, which represents the number of successes. Here, success refers to the number greater than 4. P(X = 0) = P(number less than or equal to 4 on both the tosses) = P(X = 1) = P(number less than or

Q: Question 10: Find the expected value, variance, and standard deviation of the random variable whose p.m.f’s are given below:

(i) We construct the following table to find the expected value, variance, and standard deviation: (ii) We construct the following table to find the expected value, variance, and standard deviation: (iii) We construct the following table to find the expected value, variance, and standard deviation:

Q: Question 11: A player tosses two wins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears and ₹ 2 if no head appears. Find the expected winning amount and variance of the winning amount.

When a coin is tossed twice, the sample space is S = {HH, HT, TH, HH} Let X denote the amount he wins. Then X takes values 10, 5, 2. P(X = 10) = P(2 heads appear) = P(X = 5) = P(1 head appears) = = P(X = 2) = P(no head appears) = We construct the following table to calculate the mean and the varianc

Question 1

Question 1: P.d.f. of a c.r.v. X is f(x) = 6x(1 – x), for 0 ≤ x ≤ 1 and = 0, otherwise (elsewhere) If P(X < a) = P(X > a), then a = (a) 1 (b) (c) (d)

Accepted Answer

(b)

Question 2

Question 2: If the p.d.f. of a c.r.v. X is f(x) = 3(1 – 2x2), for 0 < x < 1 and = 0, otherwise (elsewhere), then the c.d.f. of X is F(x) = (a) 2x – 3x2 (b) 3x – 4x3 (c) 3x – 2x3 (d) 2x3

Accepted Answer

(c) 3x – 2x3

Question 3

Question 3: If the p.d.f. of a c.r.v. X is f(x) = , for -3 < x < 3 and = 0, otherwise, then P(|X| < 1) = (a) (b) (c) (d)

Accepted Answer

(a)

Question 4

Question 4: If p.m.f. of a d.r.v. X takes values 0, 1, 2, 3, … which probability P(X = x) = k(x +1). 5-x, where k is a constant, then P(X = 0) = (a) (b) (c) (d)

Accepted Answer

(b)

Question 5

Question 5: If p.m.f. of a d.r.v. X is P(X = x) = , for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. If a = P(X ≤ 2) and b = P(X ≥ 3), then (a) a < b (b) a > b (c) a = b (d) a + b

Accepted Answer

(c) a = b

Question 6

Question 6: If p.m.f. of a d.r.v. X is P(X = x) = , for x = 1, 2, 3, ……, n and = 0, otherwise, then E(X) = (a) (b) (c) (d)

Accepted Answer

(b)

Question 7

Question 7: If p.m.f. of a d.r.v. X is P(x) = , for x = 1, 2, 3 and = 0, otherwise (elsewhere), then E(X) = (a) (b) (c) (d)

Accepted Answer

(b)

Question 8

Question 8: If the d.r.v. X has the following probability distribution: then P(X = -1) = (a) (b) (c) (d)

Accepted Answer

(a)

Question 9

Question 9: If the d.r.v. X has the following probability distribution: then k = (a) (b) (c) (d)

Accepted Answer

(d)

Question 10

Question 10: Find the expected value of X for the following p.m.f. (a) 0.85 (b) -0.35 (c) 0.15 (d) -0.15

Accepted Answer

(b) -0.35

Question 11

Question 1: Identify the random variable as either discrete or continuous in each of the following. If the random variable is discrete, list its possible values: (i) An economist is interested

Accepted Answer

(i) Let X = number of unemployed graduates in a town. Since the population of the town is 1 lakh, X takes the finite values. ∴ random variable X is discrete. Range = {0, 1, 2, …, 99999, 100000}.

Question 12

Question 2: The probability distribution of discrete r.v. X is as follows: (i) Determine the value of k. (ii) Find P(X ≤ 4), P(2 < X < 4), P(X ≥ 3).

Accepted Answer

Step-by-step solution provided.

Question 13

Question 3: The following is the probability distribution of X: Find the probability that (i) X is positive (ii) X is non-negative (iii) X is odd (iv) X is even.

Accepted Answer

(i) P(X is positive) = P(X = 1) + P(X = 2) + P(X = 3) = 0.25 + 0.15 + 0.1 = 0.50

Question 14

Question 4: The p.m.f. of a r.v. X is given by P(X = x) = x = , for x = 0, 1, 2, 3, 4, 5 and = 0, otherwise. Then show that P(X ≤ 2) = P(X ≥ 3).

Accepted Answer

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = = ………{ }^{n} C_{r}={ }^{n} C_{n-r} = P(X = 5) + P(X = 4) + P(X = 3) = P(X ≥ 3) ∴ P(X ≤ 2) = P(X ≥ 3).

Question 15

Question 5: In the p.m.f. of r.v. X Find a and obtain c.d.f. of X.

Accepted Answer

For p.m.f. of a r.v. X ∴ P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 1 Let F(x) be the c.d.f. of X. Then F(x) = P(X ≤ x) ∴ F(1) = P(X ≤ 1) = P(X = 1) = F(2) = P(X ≤ 2) = P(X = 1) + P (X = 2) P(3) = P(X ≤ 3) = P(X = 1) + P(X = 2) + P(X = 3) F(4) = P(X ≤ 4) = P(X = 1) + P(X = 2) + P(X = 3)

Question 16

Question 6: A fair coin is tossed 4 times. Let X denote the number of heads obtained. Write down the probability distribution of X. Also, find the formula for p.m.f. of X.

Accepted Answer

When a fair coin is tossed 4 times then the sample space is S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, HTTH, THHT, THTH, TTHH, HTTT, THTT, TTHT, TTTH, TTTT} ∴ n(S) = 16 X denotes the number of heads. ∴ X can take the value 0, 1, 2, 3, 4 When X = 0, then X = {TTTT} ∴ n (X) = 1 ∴ P(X = 0) = When X

Question 17

Question 7: Find the probability distribution of the number of successes in two tosses of a die, where success is defined as (i) number greater than 4 (ii) six appear on at least one die.

Accepted Answer

When a die is tossed two times, we obtain (6 × 6) = 36 number of observations. Let X be the random variable, which represents the number of successes. Here, success refers to the number greater than 4. P(X = 0) = P(number less than or equal to 4 on both the tosses) = P(X = 1) = P(number less than or

Question 18

Question 8: A random variable X has the following probability distribution: Determine: (i) k (ii) P(X > 6) (iii) P(0 < X < 3). Question 9. The following is the c.d.f. of a r.v. X: Fin

Accepted Answer

(i) From the given table F(-3) = 0.1, F(-2) = 0.3, F(-1) = 0.5 F(0) = 0.65, f(1) = 0.75, F(2) = 0.85 F(3) = 0.9, F(4) = 1 P(X = -3) = F(-3) = 0.1 P(X = -2) = F(-2) – F(-3) = 0.3 – 0.1 = 0.2 P(X = -1) = F(-1) – F(-2) = 0.5 – 0.3 = 0.2 P(X = 0) = F(0) – F(-1) = 0.65 – 0.5 = 0.15 P(X = 1) = F(1) – F(0)

Question 19

Question 10: Find the expected value, variance, and standard deviation of the random variable whose p.m.f’s are given below:

Accepted Answer

(i) We construct the following table to find the expected value, variance, and standard deviation: (ii) We construct the following table to find the expected value, variance, and standard deviation: (iii) We construct the following table to find the expected value, variance, and standard deviation:

Question 20

Question 11: A player tosses two wins. He wins ₹ 10 if 2 heads appear, ₹ 5 if 1 head appears and ₹ 2 if no head appears. Find the expected winning amount and variance of the winning amount.

Accepted Answer

When a coin is tossed twice, the sample space is S = {HH, HT, TH, HH} Let X denote the amount he wins. Then X takes values 10, 5, 2. P(X = 10) = P(2 heads appear) = P(X = 5) = P(1 head appears) = = P(X = 2) = P(no head appears) = We construct the following table to calculate the mean and the varianc

Question 21

Question 12: Let the p.m.f. of r.v. X be P(x) = , for x = -1, 0, 1, 2 and = 0, otherwise. Calculate E(X) and Var(X).

Accepted Answer

P(X) = X takes values -1, 0, 1, 2 P(X = -1) = P(-1) = P(X = 0) = P(0) = P(X = 1) = P(1) = P(X = 2) = P(2) = We construct the following table to calculate the mean and variance of X: From the table ΣxiP(xi) = 0 and = 1 E(X) = ΣxiP(xi) = 0 Var(X) = – [E(X)]2 = 1 – 0 = 1 Hence, E(X) = 0, Var (X) = 1.

Question 22

Question 13: Suppose the error involved in making a certain measurement is a continuous r.v. X with p.d.f. f(x) = k(4 – x2), -2 ≤ x ≤ 2 and = 0 otherwise. Compute (i) P(X > 0) (ii) P(-1 <

Accepted Answer

(i) P(X > 0) (ii) P(-1 < X < 1) (iii) P(X < -0.5 or X > 0.5)

Question 23

Question 14: The p.d.f. of a continuous r.v. X is given by f(x) = , for 0 < x < 2a and = 0, otherwise. Show that P( X < ) = P(X > )

Accepted Answer

Step-by-step solution provided.

Question 24

Question 15: The p.d.f. of r.v. X is given by f(x) = , for 0 < x < 4 and = 0, otherwise. Determine k. Determine c.d.f. of X and hence find P(X ≤ 2) and P(X ≤ 1).

Accepted Answer

Since f is p.d.f. of the r.v. X,

Question 25

Question 9: The following is the c.d.f. of a r.v. X: Find (i) p.m.f. of X (ii) P( -1 ≤ X ≤ 2) (iii) P(X ≤ X > 0).

Accepted Answer

block bg-white dark:bg-slate-900/80 border-l-4 border-emerald-500 rounded-r-xl p-5 my-3 shadow-xs border border-slate-200/60 dark:border-slate-800"> Solution & Step-by-Step Answer: (i) From the given table F(-3) = 0.1, F(-2) = 0.3, F(-1) = 0.5 F(0) = 0.65, f(1) = 0.75, F(2) = 0.85 F(3) = 0.9, F(4) =

Chapter 7 Probability Distributions Miscellaneous Exercise 7 Solutions