It is known that the amounts of time required for room-service delivery at a certain Marriott Hotel are Normally distributed with the average delivery time of 20 minutes. Answer the following questions. Question 1 (3 points). In case a random sample of 11 deliveries was selected and the standard deviation of these delivery times was found to be 7.87 minutes, find the probability that their average delivery time is less than 23 minutes.

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Answer:

64.80% probability that their average delivery time is less than 23 minutes.

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 = 20, \sigma = 7.87[/tex]

Find the probability that their average delivery time is less than 23 minutes.

This is the pvalue of Z when X = 23.

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

[tex]Z = \frac{23 - 20}{7.87}[/tex]

[tex]Z = 0.38[/tex]

[tex]Z = 0.38[/tex] has a pvalue of 0.6480

64.80% probability that their average delivery time is less than 23 minutes.