The amount of time a bank teller spends with each customer is normally distributed with a mean of 3.10 minutes and a standard
deviation of 0.40 minutes. (Use the z table)
If a random sample of 16 customers is selected from the bank, what is the probability that the average time spent per customer will be at least 3 minutes?

There is an 85 percent chance that the sample mean will be below how many minutes?

Doubting Thomas does not believe that the mean is actually 3.10 minutes since he has often waited longer. He decides to conduct a study and generates a random sample of 25 customers. He calculates the waiting time for all 25 customers and finds the average A waiting time to be 4 minutes. Based on this information, what is the best estimate of the population mean assuming 95 percent accuracy. Is the doubting Thomas correct in his assumption?

Solution Summary

The solution provides step by step method for the calculation of Normal probabilities and t test for mean. Formula for the calculation and Interpretations of the results are also included. Interactive excel sheet is included. The user can edit the inputs and obtain the complete results for a new set of data.

... 2d. Normal Probabilities. ...Probability for X > X Value 106 Z Value 2.711088342 P(X>106) 0.0034. Page 1 Qn. 2c. t Test for Hypothesis of the Mean. ...

... by step method for the calculation of normal probability, testing of hypothesis and ...Normal Probabilities. ... random samples were taken from normal distributions of ...

... the Mean 1.0000 Degrees of Freedom 24 t Test Statistic -0.8000. Two-Tail Test Lower Critical Value -2.063898547 Upper Critical ...Normal Probability Distribution. ...

...Normal Probabilities. Common Data Mean 62 Standard Deviation 6 Probability for a Range ... the Mean 12.64911064 Degrees of Freedom 39 t Test Statistic -1.58113883. ...

...Normal Probabilities. ...Probability for X<6 or X >7 P(X<6 or X >7 ... 64871111.11 Standard Error 3620.1209 Difference in Sample Means 5300 t Test Statistic 1.464039522. ...

... Standard Error of the Mean 2.828427125 Degrees of Freedom 49 t Value 2.009575237 ... Normal Probabilities. ...Probability for X > X Value 66000 Z Value 3 P(X>66000 ...

... Standard Error of the Mean 4.546060566 Degrees of Freedom 6 t Value 2.446911851 ...Normal Probabilities. ...Probability for X <= X Value 20 Z Value 0.2731707 P(X<=20 ...

... Thus at alpha = 0.01, the probability of making a ... sample is selected from a normal population distribution ... standard deviation we will use t distribution sample ...

... 3, use the formula to find the following probabilities. ... z value and then we find out the probability. ... 13.8 Random sampling from two normal populations produced ...