The amount of time a bank teller spends with each customer has a population
mean of 3.10 minutes and standard deviation of 0.40 minute. If a random sample
of 16 customers is selected,
a) what assumption must be made in order to solve parts (b) and (c)?
b) what is the probability that the average time spent per customer will be at
least 3 minutes?
c) there is an 85% chance that the sample mean will be below how many
minutes?
d) If a random sample of 64 customers is selected, there is an 85% chance that
the sample mean will be below how many minutes?
The manager of Lone Star Restaurant believes that the average wait time of the Saturday evening meal is 30 minutes. To test this belief, the manager selects 50 customers at random and computes the average wait time to be 34 minutes. Assume a population standard deviation of 12 minutes.
a) In order to solve parts (b) and (c), the assumption that must be made is that the sampling distribution of the sample mean follows a normal distribution.
b) To find the probability that the average time spent per customer will be at least 3 minutes, we need to calculate the probability of the sample mean being greater than or equal to 3 minutes.
First, we need to standardize the value by calculating the z-score:
z = (x - μ) / (σ / √n)
where x is the value we want to find the probability for, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x = 3 minutes, μ = 3.10 minutes, σ = 0.40 minutes, and n = 16 customers.
Substituting the values, we get:
z = (3 - 3.10) / (0.40 / √16) = -0.10 / (0.40 / 4) = -0.10 / 0.10 = -1
Next, we use a standard normal distribution table or a calculator to find the probability that the z-score is greater than or equal to -1. From the table, we can see that the probability of a z-score being greater than or equal to -1 is approximately 0.8413.
Therefore, the probability that the average time spent per customer will be at least 3 minutes is 0.8413 or 84.13%.
c) To find the number of minutes below which there is an 85% chance that the sample mean will be, we need to find the z-score corresponding to the 85th percentile.
Using the standard normal distribution table or a calculator, we look for the z-score that gives an area of 0.85 (85%) to the left.
From the table, we find that the z-score corresponding to an area of 0.85 is approximately 1.036.
We can use the z-score formula to find the sample mean:
z = (x - μ) / (σ / √n)
Rearranging the formula, we have:
x = μ + (z * σ / √n)
Substituting the values, we get:
x = 3.10 + (1.036 * 0.40 / √16) = 3.10 + (1.036 * 0.40 / 4) = 3.10 + (0.1036) = 3.20
Therefore, there is an 85% chance that the sample mean will be below 3.20 minutes.
d) Similar to part (c), to find the number of minutes below which there is an 85% chance that the sample mean will be, we follow the same steps.
Using the standard normal distribution table or a calculator, we find that the z-score corresponding to an area of 0.85 is approximately 1.036.
Again, we use the z-score formula:
x = μ + (z * σ / √n)
Substituting the values, we get:
x = 3.10 + (1.036 * 0.40 / √64) = 3.10 + (1.036 * 0.40 / 8) = 3.10 + 0.0518 = 3.1518
Therefore, there is an 85% chance that the sample mean will be below 3.1518 minutes.