Question

The speed with which utility companies can resolve problems is very important. GTC, the Georgetown Telephone Company, reports it can resolve customer problems the same day they are reported in 70% of the cases. Suppose the 12 cases reported today are representative of all complaints.

a-1. How many of the problems would you expect to be resolved today? (Round your answer to 2 decimal places.)

a-2. What is the standard deviation? (Round your answer to 4 decimal places.)

b. What is the probability 6 of the problems can be resolved today? (Round your answer to 4 decimal places.)

c. What is the probability 6 or 7 of the problems can be resolved today? (Round your answer to 4 decimal places.)

d. What is the probability more than 7 of the problems can be resolved today? (Round your answer to 4 decimal places.)

Answer 1

Step-by-step explanation:

a-1. We can use the expected value formula to find the number of problems we would expect to be resolved today:

Expected value = Number of cases reported x Probability of resolution

Expected value = 12 x 0.7

Expected value = 8.4

Rounding to 2 decimal places, we would expect 8.40 problems to be resolved today.

a-2. To find the standard deviation, we can use the formula:

Standard deviation = √(n x p x (1 - p))

where n is the number of cases reported and p is the probability of resolution. Substituting in the values given, we get:

Standard deviation = √(12 x 0.7 x (1 - 0.7))

Standard deviation = √(3.36)

Standard deviation = 1.8326

Rounding to 4 decimal places, the standard deviation is 1.8326.

b. To find the probability that 6 of the problems can be resolved today, we can use the binomial probability formula:

P(X = x) = (nCx) * p^x * (1 - p)^(n - x)

where n is the number of cases reported, p is the probability of resolution, x is the number of cases that can be resolved, and nCx is the binomial coefficient (nCx = n! / x!(n-x)!).

Substituting in the values given, we get:

P(X = 6) = (12C6) * 0.7^6 * (1 - 0.7)^(12 - 6)

P(X = 6) = (924) * 0.7^6 * 0.3^6

P(X = 6) = 0.0482

Rounding to 4 decimal places, the probability that 6 of the problems can be resolved today is 0.0482.

c. To find the probability that 6 or 7 of the problems can be resolved today, we can use the same formula as in part (b), but we need to calculate the probabilities separately for 6 and 7, and then add them together:

P(X = 6 or X = 7) = P(X = 6) + P(X = 7)

P(X = 6 or X = 7) = (12C6) * 0.7^6 * 0.3^6 + (12C7) * 0.7^7 * 0.3^5

P(X = 6 or X = 7) = 0.0482 + 0.1254

P(X = 6 or X = 7) = 0.1736

Rounding to 4 decimal places, the probability that 6 or 7 of the problems can be resolved today is 0.1736.

d. To find the probability that more than 7 of the problems can be resolved today, we can use the complement rule and subtract the probability of 7 or fewer problems being resolved from 1:

P(X > 7) = 1 - P(X ≤ 7)

P(X > 7) = 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) + P(X = 6) + P(X = 7)]

P(X > 7) = 1 - [(12C0) * 0.7^0 * 0.3^12 + (12C1) * 0.7