Assume that the heights of adult caucasian women have a mean of 63.6 inches and a standard deviation of 2.5 inches. if 100 women are randomly selected, find the probability that they have a mean height greater than 63.0 inches. round to four decimal places. a.0.9918b.0.8989c.0.2881d.0.0082e. not enough information to determine

A. 0.9918

Step-by-step explanation:

Mean Height = u = 63.6 inches

Standard deviation = = 2.5 inches

Sample size = 100

We have to find the probability that the mean height of sample would be greater than 63 i.e. P(x > 63)

We can do this by converting this value to z-score.

The formula to find z-score given the mean, standard deviation and sample size is:

Using the values, we get:

So now we have to find the probability that the z score would be greater than -2.4

i.e. P(z > -2.4)

From the z-table this value comes out to be:

P(z > -2.4) = 0.9918

Since,

P ( x > 63.0) is equivalent to P(z > -2.4), we can say:

P( x > 63.0 ) = 0.9918

Thus, the probability that adult Caucasian women have a mean height greater than 63.0 inches is 0.9918

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