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The weight of a single bag checked by an airplane passenger follows a distribution that is right skewed with a mean of 38 pounds and a standard deviation of 6.2 pounds. If a random sample of 96 bags is selected, what is the probability that the average weight of the bags exceeds 40 pounds?

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Answer:

0.0007872

Step-by-step explanation:

-Since the sample size is large enough, we apply the normal distribution to find the probability.

#Given n=96, mean=38 and standard deviation=6.2 pounds, the probability can be calculated as:

[tex]P(X>40)=P(z>\frac{X-\mu}{\sigma/\sqrt{n}})\\\\=P(z>\frac{40-38}{6.2/\sqrt{96}})\\\\=P(z>3.1606}\\\\=0.0007872[/tex]

Hence, the probability is 0.0007872

Answer:

0.0008

Step-by-step explanation:

P(z>3.16) = 0.0008.

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