Service times for customers at a post office follow some right-skewed distribution with mean 2.91 minutes and standard deviation 1.74 minutes.

(a) Can you calculate the probability that the average service time for the next two customers is less than 2.64 minutes? If you can, enter the probability in the answer box. If you can't, type the word No.

(b) What is the probability that the mean service time for a random sample of 35 customers is between 2.53 and 3.04 minutes?

(c) What is the probability that the total service time for the next 41 customers exceeds 134.02 minutes? (Assume that the next 41 customers represent a simple random sample and that there is always a customer waiting in line.)

(d) What is the probability that the mean service time for a random sample of 54 customers is exactly 3.59 minutes?

To answer these questions, we will use the properties of the normal distribution. Given that the sample size is sufficiently large (n > 30) and under the assumption that the sample is representative of the population, we can use the Central Limit Theorem to approximate the distribution of the sample mean as normal.

(a) To calculate the probability that the average service time for the next two customers is less than 2.64 minutes, we need to find the probability that the sample mean is less than 2.64 minutes. Since the population mean is 2.91 minutes and the standard deviation is 1.74 minutes, we compute the standard error of the sample mean as follows:

Standard Error = standard deviation / sqrt(sample size)
= 1.74 / sqrt(2)
≈ 1.23 minutes

Next, we standardize the sample mean using the formula:

Z = (x - mean) / standard error

where x is the sample mean. Substituting the values, we find:

Z = (2.64 - 2.91) / 1.23
≈ -0.22

Now, we can use a standard normal distribution table or a calculator to find the probability associated with Z = -0.22. In this case, the probability is approximately 0.413. Therefore, the probability that the average service time for the next two customers is less than 2.64 minutes is approximately 0.413.

(b) To find the probability that the mean service time for a random sample of 35 customers is between 2.53 and 3.04 minutes, we once again need to standardize the sample mean. The standard error for this sample size is given by:

Standard Error = standard deviation / sqrt(sample size)
= 1.74 / sqrt(35)
≈ 0.294 minutes

Standardizing the lower and upper bounds of the desired range, we get:

Z(lower) = (2.53 - 2.91) / 0.294
≈ -1.29

Z(upper) = (3.04 - 2.91) / 0.294
≈ 0.44

Using the standard normal distribution table or a calculator, we can find the probabilities associated with Z(lower) and Z(upper). The probability associated with Z(lower) is approximately 0.098 and the probability associated with Z(upper) is approximately 0.670. Subtracting these two probabilities, we get the probability that the mean service time for a random sample of 35 customers is between 2.53 and 3.04 minutes, which is approximately 0.572.

(c) To find the probability that the total service time for the next 41 customers exceeds 134.02 minutes, we need to consider the distribution of the sum of service times rather than the mean. In this case, the sum of service times follows a normal distribution with mean equal to the population mean multiplied by the sample size (2.91 * 41 = 119.31 minutes) and standard deviation equal to the population standard deviation multiplied by the square root of the sample size (1.74 * sqrt(41) ≈ 9.61 minutes).

Now, we need to find the probability that the sum of service times is greater than 134.02 minutes. To do this, we can standardize the value using the formula:

Z = (x - mean) / standard deviation

where x is the value we want to find the probability for. Substituting the values, we get:

Z = (134.02 - 119.31) / 9.61
≈ 1.53

Looking up the probability associated with Z = 1.53 in the standard normal distribution table or using a calculator, we find that the probability is approximately 0.937. Therefore, the probability that the total service time for the next 41 customers exceeds 134.02 minutes is approximately 0.937.

(d) To find the probability that the mean service time for a random sample of 54 customers is exactly 3.59 minutes, we need to account for the fact that the mean is a continuous variable and has an infinite number of possible values. Therefore, the probability of the sample mean being exactly equal to 3.59 minutes is infinitesimally small. Instead, we can consider the probability of the sample mean falling within a small range around 3.59 minutes.

If we choose a small range around 3.59 minutes, we can use the techniques described in part (b) to compute the probability of the sample mean falling within that range. The range could be, for example, 3.58 to 3.60 minutes. By calculating the associated probabilities, we would get an approximation of the probability that the mean service time for a random sample of 54 customers falls within this small range.