After a repeated observation, it has been determined that the waiting time at the drive through window of a local bank is skewed left, with a mean of 3.5 minutes and a standard deviation of 1.9 minutes. a random sample of 100 customers is to be taken. What is the probability that the mean of the sample will exceed 4 minutes? show calculation
After a repeated observation, it has been determined that the waiting time at the drive through...
The amount of time that a drive-through bank teller spends on a customer is a random variable with a mean 3.2 minutes and a standard deviation a = 1.6 minutes. If a random sample of 64 customers is observed, find the probability that their mean time at the teller's window is A. at most 2.7 minutes; B. more than 3.5 minutes; C. at least 3.2 minutes but less than 3.4 minutes (10 pts. each, 30 pts. total)
1. At a fast food restaurant, the waiting time at the drive-through window has an average of 3 minutes, with a standard deviation of 0.8 minutes. i. What is the probability that a random sample of 64 cars will have an average waiting time of less than 3.25 minutes? ii. Did you use the CLT to do this problem? Explain.
A random sample of 64 customers at a drive-through bank window is observed, and it is found that the teller spends an average of 2.8 minutes with each customer, with a standard deviation of 1.2 minutes. Find a 93% confidence interval for the true mean time that this teller takes with her customers.
The amount of time that a drive-through bank teller spends on a customer is a random variable with a mean u = 7.9 minutes and a standard deviation o = 3.6 minutes. If a random sample of 81 customers is observed, find the probability that their mean time at the teller's window is (a) at most 7.3 minutes; (b) more than 8.7 minutes; (c) at least 7.9 minutes but less than 8.3 minutes. Click here to view page 1 of...
The waiting time until a customer is served at a fast food restaurant during lunch hours has a skewed distribution with a mean of 2.4 minutes and a standard deviation of 0.4 minute. Suppose that a random sample of 44 waiting times will be taken. Compute the probability that the mean waiting time for the sample will be longer than 2.5 minutes. Answer: (Round to 4 decimal places.)
A local hardware store claims that the mean waiting time in line is less than 3.5 minutes. A random sample of 20 customers has a mean of 3.3 minutes with a standard deviation of 0.8 minute. If a = 0.05, test the store's claim. Assumption: ? Parameter: ? Hypothesis: ? Test Statistic: ? Reject- Region: ? Calculated Test Statistic: ? Conclusion: ? P-value: ?
d) To test if the mean waiting time at the drive-through window at a fast food restaurant during rush hour differs from 10 minutes. Hypothesis to be tested (using symbols): H0 : H1 :
You are the manager of a restaurant for a fast-food franchise.
Last month, the mean waiting time at the drive-through window for
branches in your geographical region, as measured from the time a
customer places an order until the time the customer receives the
order, was 3.8 minutes. You select a random sample of 81 orders.
The sample mean waiting time is 3.63 minutes, with a sample
standard deviation of 0.9 minute. Complete parts (a) and (b)
below.
fast-food franchise....
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of n-49 independent customers is observed. Find the approximate probability that the average time waiting in line for these customers is (a) Greater than 10 minutes (b) Between 6 and 10 minutes (c) Less than 6 minutes
A change store claims that the mean of waiting time is 3.5 minutes. A regular customer who happened to be taking a statistics and probability class believes that the wait time is longer. The student then gathers a random sample of 20 customers which has a mean of 3.8 minutes with a standard deviation of 0.5 minute. If alpha=0.05, test the store's claim. Ho: H1: Test statistic: Critical value: P value: Conclusion: