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Fast-food restaurants spend much time studying the amount of time cars spend in their drive-thrus. Certainly, the faster the cars get service, the more opportunity for making money. According to a recent study by QSR magazine, Wendy’s has the best time, with a mean time spent in the drive thru of 138.5 seconds. Assuming drive-thru time is normally distributed with a standard deviation of 29 seconds, what proportion of cars spends between 120 and 180 seconds in Wendy's drive-thru? Write answer as decimal rounded to the thousandth.

Respuesta :

Answer:

0.663

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

[tex]\mu = 138.5, \sigma = 29[/tex]

What proportion of cars spends between 120 and 180 seconds in Wendy's drive-thru?

This is the pvalue of Z when X = 180 subtracted by the pvalue of Z when X = 120. So

X = 180

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{180 - 138.5}{29}[/tex]

[tex]Z = 1.43[/tex]

[tex]Z = 1.43[/tex] has a pvalue of 0.924

X = 120

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{120 - 138.5}{29}[/tex]

[tex]Z = -0.64[/tex]

[tex]Z = -0.64[/tex] has a pvalue of 0.261

0.924 - 0.261 = 0.663

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