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A particular fruit's weights are normally distributed, with a mean of 794 grams and a standard...

A particular fruit's weights are normally distributed, with a mean of 794 grams and a standard deviation of 5 grams.

If you pick 31 fruits at random, then 16% of the time, their mean weight will be greater than how many grams?

Give your answer to the nearest gram.

math   Statistics-And-Probability   
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Answer #1

Mean = \mu = 794

Standard deviation = \sigma = 5

Sample size = n = 31

We have given P(X > \bar{x}) = 0.16

z value 0.16 is 0.99

We have to find value of \bar{x}

\bar{x}=\mu+z*\sigma/\sqrt{n}=794+0.99*5/\sqrt{31}=794+0.8890=794.889

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