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The owner of a fish market has an assistant who has determined that the weights of catfish are normally​ distributed, with mean of 3.2 pounds and standard deviation of 0.8 pound. If a sample of 64 fish yields a mean of 3.4​ pounds, what is probability of obtaining a sample mean this large or​ larger? Round to four decimal places.

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Answer:

[tex]P(X\geq 3.4)=0.0228[/tex]

Step-by-step explanation:

Given the mean is 3.2, standard deviation is 0.8 and the sample size is 64.

-We calculate the  probability of a mean of 3.4 as follows:

#First determine the z-value:

[tex]z=\frac{\bar X -\mu}{\sigma/\sqrt{n}}\\\\=\frac{3.4-3.2}{0.8/\sqrt{64}}\\\\=2.000[/tex]

#We then determine the corresponding probability on the z tables:

[tex]Z(X\geq 3.4)=1-P(X<3.4)\\\\=1-0.97725\\\\=0.0228[/tex]

Hence, the probability of obtaining a sample mean this large or​ larger is 0.0228

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