Probability Distribution Mean (Expectation), Variance :: Problems

No. Problems & Solutions [Click Hide/Show to display the solutions below the question]
01. Find the mean and variance of the following probability distribution and hence find its standard deviation
x − 10 − 20 30 75 80
P (X = x)
 1 5
 3 20
 1 2
 1 10
 1 20

Solution
[Expectation: 21.5; Variance: 932.25 ; Standard Deviation : +30.53]

02. A bakery has the following schedule of daily demand for cakes.
 No. of cakes demanded (in hundreds) Probability 0 1 2 3 4 5 6 7 8 9 0.02 0.07 0.09 0.12 0.2 0.2 0.18 0.1 0.01 0.01
Find the expected number of cakes demanded per day

Solution
[[Expectation: 436; Variance: 3.5304 ; Standard Deviation : +1.879]]

03. A random variable x has the following probability distribution
 Values of x P(X = x) 0 1 2 3 4 5 6 7 0 2k 2k k 3k k2 2K2 7k2 + k
Find the value of K and then evaluate P(x < 6), P(x ≥ 6), and P(0 < x < 5). Find also the mean and variance of the distribution

Solution
[Expectation: 3.46; Variance: 4.0284 ; Standard Deviation : +2.007]

04. The monthly demand for radios is known to have the following probability distribution
 Demand Probability 1 2 3 4 5 6 0.1 0.15 0.2 0.25 0.2 0.1
Determine the expected demand for radios. Also find the variance. If the cost (Rs. C) of producing (n) radios is given by C = 1000 + 200n, determine the expected cost.

Solution
[Expectation Cost: 1,720]

05. For what value of "a" will the function f(x) = ax; x = 1, 2, ..., n be the probability mass function of a discrete random variable x? find the mean and the variance of x.

Solution
[ Variance: 3.5304 ; Standard Deviation : +1.879]

06. Evaluate K if f(x) = K, x = 1, 2, 3, 4, 5, 6; f(x) = 0, elsewhere is a probability mass function. Also find its mean and standard deviation.

Solution
[Expectation: 3.5 ; Variance: 2.917 ; Standard Deviation : +1.708]

07.
 The p.m.f. of a random variable (x) is given by f(x) = 2p for x = 1, f(x) = p for x = 2, f(x) = 4p for x = 3, f(x) = 0 otherwise
Find the value of P, E(x) and Var(x). Also calculate P(X > 1).

Solution
[Expectation: 2.29 ; Variance: 0.7559 ; Standard Deviation : +0.869 ; P(X>1):0.714]

08. The range of a random variable X is (0, 1, 2). If C is a constant such that P(X) = 3C3, P(X = 1) = 4C − 10C2,
P(x = 2) = 5C − 1, then C =

Solution
[Expectation: 2.56 ; Variance: 0.4464 ; Standard Deviation : +0.668]

09. The value of C for which P(X = K) = CK2 can serve as the probability function of a random variable x that takes values 0, 1, 2, 3, 4, is ...

Solution
[Expectation: 3.33 ; Variance: 0.71 ; Standard Deviation : +0.843]

10. A random variable x takes the values of &m inus;1, 0, 1. Its mean is 0.6. If P(X = 0) = 0.2, then P(X = 1) = ?

Solution
[Expectation: 0.6 ; Variance: 0.44 ; Standard Deviation : +0.663; P(X=1): 0.7]

11. If x is a random variable, show that (i) E(1) = 1; (ii) E(3x) = 3E(x) and (iii) E(2 + 3x) = 2 + 3E(x)

Solution
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12. Prove that: (i) var(2) = 0, (ii) var(3x) = 9 (iii) var (x + 4x) = 16 var (x)

Solution
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13. If the variance of a random variable x is 5, then the variance of the random variable (− 3x) is

Solution
[Var(-3x): 45]

14. A number is chosen at random from the set 1, 2, 3, ... , 100 and another number is chosen at random from the set 1, 2, 3, ... , 50. What is the expected value of the sum and the expected value of the product?

Solution
[Expectation: 25.5 ; Expectation(sum): 76 ; Expectation(product): 1,287.75]

15. The probability that there is at least one error is an accounts statement prepared by A is 0.2 and for B and C they are 0.25 and 0.4 respectively A, B, and C prepared 10, 16 and 20 statements respectively. Find the expected number of correct statements in all

Solution
Expected no. of correct statements in all prepared by "A", "B" and "C" together : 32

No. Problems for Practice
01. If it rains, an umbrella sales man can earn Rs. 300 per day. If it is fair he can lose Rs. 60 per day. What is his expectation if the probability of rain is 0.3?