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Probability Assignments using Statistics Assignment Solution


Discrete Random Variable Assignments

1. The volunteer fire department handles between 0 to 5 calls on any given day. The probability distribution for the number of service calls is

Number of CallsProbability
0.20
1.30
2.20
3.10
4.10
5.10

a. Is this a valid probably assignment? Explain.

Yes, this is a valid probability assignment, as the sum of probabilities of all events, equals 1

b. What is the random variable in this problem?

The random variable in this problem is the number of calls.

c. What is the probability that the fire department handles at least 3 service calls on a given day?

P (Number of calls ≥ 3) = Σ P(X = x), where x varies from 3 to 5

P (Number of calls ≥ 3) = 0.1 + 0.1 + 0.1 = 0.3

d. What is the probability that the fire department handles at most 2 service calls on a given day?

P (Number of calls ≤ 2) = Σ P(X = x), where x varies from 0 to 2

P (Number of calls ≤ 2) = 0.2 + 0.3+ 0.2 = 0.

e. What is the expected number of service calls?

Expected number of calls = Σx*P(X =x) = 1*0.3 + 2*0.2 + 3*0.1 + 4*0.1 + 5*0.1 = 1.9

f. What is the variance in the number of service calls?

Variance = Σx2*p(x) – μ2

Variance = 1*0.3 + 4*0.2 + 9*0.1 + 16*0.1 + 25*0.1 – 1.9*1.9 = 2.49

g. What is the standard deviation?

Standard deviation = sqrt (Variance) = sqrt (2.49) = 1.578

2. Seven plants are operated by a garment manufacturer. They feel there is a ten percent chance for a strike at any one plant and the risk of a strike at one plant is independent of the risk of a strike at another plant.

a. What is the random variable in this problem?

Ans: Number of strikes on the seven plants.

b. What is

 1. What is defined as a “success”?

 Ans: Here “success” is a strike at a plant.

 2. What is the probability of success?

 Ans: P(Success) = 0.1

 3. What is defined as a “failure”?

 Ans: Here “failure” is not a strike at a plant.

 4. Calculate the probability of failure.

 Ans: P(Failure) = 0.9

c. Write the probability density function.

Ans: Here X follows a Binomial distribution with n=7 and p=0.1. Thus the probability density function is given by: f(x) = (7¦x) 〖0.1〗^x 〖0.9〗^(7-x), for x=0,1,..,7

d. What is the probability at most three plants will strike?

Ans: P(at most three plants will strike) = P(X≤3) =P(X=0) + P(X=1) + P(X=2) + P(X=3)

=f(0) + f(1) + f(2) + f(3) = 0.4783 + 0.372 + 0.124 + 0.023 = 0.9973

e. What is the probability that none of the plants will strike?

Ans: P(X=0) = f(0) = 0.4783

f. What is the probability that two of the plants will strike?

Ans: P(X=2) = f(2) = 0.124

g. What is the probability that at least one of the plants will strike?

Ans: P(X≥1) = 1- P(X=0) = 1-0.4783 = 0.5217

h. What is the probability that between four and six (inclusive) of the plants will strike?

Ans: P(X=4) + P(X=5) + P(X=6) = f(4)+f(5)+f(6) = 0.00255 + 0.00017 + 0.00001 = 0.00273

i. What is the expected number of plants that will strike?

Ans: The expected number of plants that will strike = np = 7*0.1 = 0.7

j. What is the variance in the number of plants that will strike?

Ans: The variance is given by: np(1-p) 7*0.1*0.9 = 0.63

k. What is the standard deviation of the number of plants that will strike?

Ans: Standard deviation is the square root of variance = √0.63=0.7937