Question

The average waiting time at a drive-in window of a local bank is 10. 3 minutes, with a standard deviation of 2. 7 minutes. Assume the variable is normally distributed. If a customer arrives at the bank, find the probability that the customer will have to wait between 4 and 9 minutes.

Answer 1

29.78% of customers will most likely probability be waiting between 4 and 9 minutes.

Explain the term z-score of the normal distribution?

• An observation's Z score indicates however many standard deviations it deviates from the mean.
• The standard normal distribution's mean is zero.
• Positive Z scores are those above the mean, while negative Z scores are those below the mean.

The formula for the z score is-

z = (x - μ)/σ

In which,

μ =  mean of 10. 3 minutes

σ = standard deviation of 2. 7 minutes.

z score for the x = 4 minutes.

z = (4 - 10.3)/2.7

z = -2.33

z score for the x = 9 minutes.

z = (9 - 10.3)/2.7

z = -0.48

Thus,

Probability that the customer will have to wait between 4 and 9 minutes.

P(4 < x < 9) = P(-0.48 < z <  -2.33)

Use z score table-

P(4 < x < 9) =  0.3085 - 0.0107

P(4 < x < 9) = 0.2978

P(4 < x < 9) = 29.78%

Thus, 29.78% of customers will most likely be waiting between 4 and 9 minutes.

To know more about the z-score of the normal distribution, here

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