Needing help with answering these two problems; thank you! Assume that women\'s

Question

Needing help with answering these two problems; thank you!

Assume that women's heights are normally distributed with a mean given by -63.3 in, and a standard deviation given by -28 in (a) If 1 woman is randomly selected, find the probability that her height is less than 64 in. (b) If 38 women are randomly selected, find the probability that they have a mean height less than 64 in (a) The probability is approximately (Round to four decimal places as needed.) (b) The probability is approximately L Round to four decimal places as needed.)

Explanation / Answer

a).

z value for 64, z = (64-63.3)/2.8 = 0.25

P( x <64) = P( z < 0.25) =0.5987

b)

standard error =2.8/sqrt(38) =0.4542

z value for 64, z = (64-63.3)/0.4542 = 1.54

P(mean x <64) = P( z < 1.54) = 0.9382

a).

Z value for 0.8544, z =(0.8544-0.8617)/0.0515 = -0.14

P( x >0.8544) =P( z >-0.14) =0.5557

b).

standard error =0.0515/sqrt(444) =0.0024

Z value for 0.8544, z =(0.8544-0.8617)/0.0024 = -3.04

P( mean x >0.8544) =P( z >-3.04) =0.9988

c).

Yes, because the probability of getting a sample mean of 0.8544 or greater when 444 candies are selected is not exceptionally small.

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