**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 Calls | Probability |

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
**