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A manufacturer knows that their items have a lengths that are skewed right, with a mean of 17.6 inches, andstandard deviation of 3.3 inches.If 37 items are chosen at random, what is the probability that their mean length is greater than 18.9 inches?(Round answer to four decimal places)Question Help: D VideoSubmit Question

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Answer:

The probability is 0.9999

Explanation:

Given that the mean is 17.6 inches

standard deviation is 3.3/37 = 0.089

We have:

z(18.9)

[tex]z=\frac{x-\mu}{\sigma}[/tex][tex]\frac{18.9-17.6}{0.089}=14.6067[/tex]

Now, we have

P(x > 18.9) = P(z > 14.6067)

= 0.9999

z(0.394) = 0.6532

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