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The average cholesterol content of a certain canned goods is 215 milligrams, and the standard deviation is 15 milligrams. Assume that the variable is normally distributed. If a sample of 25 canned goods is selected, what is the probability that the mean of the sample will be greater than 220 milligrams?​.

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VINEPLAZA07IN

Solution :

[tex]n \: (215. \frac{15}{ \sqrt{25} } = n(215 \: .3 \: )[/tex]

let's call X our sample mean.

[tex]pc \times > 220 > = - \frac{220 - 215}{ \sqrt{15} } = 0.02[/tex]

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