The probability distribution of number of calls coming to a call center every 15 minute period is shown
below.
x 5 6 7 8 9 10 11 12 13
p ( x ) 0.02 0.04 0.07 0.08 0.09 0.11 0.12 0.13 0.10
Continued…
x 14 15 16 17 18 19 20
p ( x ) 0.07 0.06 0.04 0.03 0.02 0.01 0.01
1. This probability distribution is an example of a
(a) continuous probability distribution
(b) discrete probability distribution
(c) binomial distribution
(d) Poisson distribution
(e) None of the above
2. The probability of no more than 10 calls in a 15 minute interval:
(a) 0.55
(b) 0.60
(c) 0.30
(d) 0.35
(e) 0.41
3. The probability of getting at least 8 calls in a 15 minute interval:
(a) 0.85
(b) 0.79
(c) 0.13
(d) 0.87
(e) 0.89
4. The expected number of calls is:
(a) 12.24
(b) 13.51
(c) 11.34
(d) 0.54
(e) none of the above
5. The standard deviation of the calls are
(a) 1.54
(b) 1.45
(c) 4.35
(d) 3.25
(e) 2.36
1. To determine the type of probability distribution, we need to assess if the random variable is discrete or continuous. In this case, the random variable represents the number of calls, which is a discrete variable. Therefore, the probability distribution is a discrete probability distribution. Therefore, the answer is (b) discrete probability distribution.
2. To find the probability of no more than 10 calls in a 15-minute interval, we sum the probabilities for x values less than or equal to 10.
P(X ≤ 10) = P(X = 5) + P(X = 6) + P(X = 7) + P(X = 8) + P(X = 9) + P(X = 10)
P(X ≤ 10) = 0.02 + 0.04 + 0.07 + 0.08 + 0.09 + 0.11
P(X ≤ 10) = 0.41
Therefore, the answer is (e) 0.41.
3. To find the probability of getting at least 8 calls in a 15-minute interval, we sum the probabilities for x values greater than or equal to 8.
P(X ≥ 8) = P(X = 8) + P(X = 9) + P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18) + P(X = 19) + P(X = 20)
P(X ≥ 8) = 0.08 + 0.09 + 0.11 + 0.12 + 0.13 + 0.10 + 0.07 + 0.06 + 0.04 + 0.03 + 0.02 + 0.01 + 0.01
P(X ≥ 8) = 0.85
Therefore, the answer is (a) 0.85.
4. The expected number of calls can be calculated by multiplying each x value by its respective probability and summing them up.
Expected number of calls = (5 * 0.02) + (6 * 0.04) + (7 * 0.07) + (8 * 0.08) + (9 * 0.09) + (10 * 0.11) + (11 * 0.12) + (12 * 0.13) + (13 * 0.10) + (14 * 0.07) + (15 * 0.06) + (16 * 0.04) + (17 * 0.03) + (18 * 0.02) + (19 * 0.01) + (20 * 0.01)
Expected number of calls = 11.34
Therefore, the answer is (c) 11.34.
5. The standard deviation of the calls can be calculated using the formula:
Standard deviation = sqrt(Σ(x - μ)^2 * p(x))
where μ is the expected value calculated in the previous question.
Standard deviation = sqrt((5 - 11.34)^2 * 0.02 + (6 - 11.34)^2 * 0.04 + ... + (20 - 11.34)^2 * 0.01)
Standard deviation ≈ 3.25
Therefore, the answer is (d) 3.25.