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Suppose a certain airline uses passenger seats that are 16.2 inches wide. Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. If an airliner is filled with 110 randomly selected adult men, what is the probability that any one of those adult male will have a hip width greater than 16.2 inches? What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches? Round your answers to three decimal places; add trailing zeros as needed. The probability that an individual adult male passenger will have an average hip width greater than 16.2 inches is . The probability that 110 randomly selected adult male passengers will have an average hip width greater than 16.2 inches is .

Respuesta :

Answer:

Each adult male has a 5.05% probability of having a hip width greater than 16.2 inches.

There is a 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

Step-by-step explanation:

Problems of normally distributed samples can be 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]

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem

Assume that adult men have hip breadths that are normally distributed with a mean of 14.4 inches and a standard deviation of 1.1 inches. This means that [tex]\mu = 14.4, \sigma = 1.1[/tex].

What is the probability that any one of those adult male will have a hip width greater than 16.2 inches?

For each one of these adult males, the probability that they have a hip width greater than 16.2 inches is 1 subtracted by the pvalue of Z when X = 16.2. So:

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

[tex]Z = \frac{16.2 - 14.4}{1.1}[/tex]

[tex]Z = 1.64[/tex]

[tex]Z = 1.64[/tex] has a pvalue of 0.9495.

This means that each male has a 1-0.9495 = 0.0505 = 5.05% probability of having a hip width greater than 16.2 inches.

For the average of the sample

What is the probability that the 110 adult men will have an average hip width greater than 16.2 inches?

Now, we need to find the standard deviation of the sample before using the zscore formula. That is:

[tex]s = \frac{\sigma}{\sqrt{110}} = 0.1[/tex].

Now

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

[tex]Z = \frac{16.2 - 14.4}{0.1}[/tex]

[tex]Z = 18[/tex]

[tex]Z = 18[/tex] has a pvalue of 0.9999.

This means that there is a 1-0.9999 = 0.0001 = 0.01% probability that the 110 adult men will have an average hip width greater than 16.2 inches.

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