2. The distribution of the amount of time customers must wait for service at the

Question

2. The distribution of the amount of time customers must wait for service at the drive-thru window of the Marysville Bank of America is well-modeled by the Normal Distribution with mean 9.2 minutes and standard deviation 2.6 minutes. Determine the percentage of all this bank's drive-thru customers that wait less than 8 minutes for service at the drive-thru window. Report this percentage with at least three digits after the decimal point. Please use your (TI-84) calculator (and not a table of values) to determine the requested percentage. Don't forget to state the name of the calculator function used (do not, however, describe every button pushed). Also include the values you had to input into this function in order to produce an output.

Explanation / Answer

Solution:

Here, we have to find P(X<8)

(By using Z-score)

We are given X follows normal distribution with mean = 9.2 and SD = 2.6

Z = (X – mean) / SD = (8 – 9.2) / 2.6 = -1.2/2.6 = -0.46154

P(Z< -0.46154) = P(X<8) = 0.322206

Required probability = 0.322206

(By using Ti-84 calculator)

Press Start > Press DISTR > Select normalcdf( > input 1E-99, 8, 9.2, 2.6) > Press Okay/Enter

Output:

.3222061669

Required percentage = 32.221%

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