Suppose that the mean value of interpupillary distance (the distance between the

Question

Suppose that the mean value of interpupillary distance (the distance between the pupils of the left and right eyes) for adult males is 65 mm and that the population standard deviation is 5 mm. (a) If the distribution of interpupillary distance is normal and a sample of n =25 adult males is to be selected, what is the probability that the sample average distance for these 25 will be between 64 and 66 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.) P = ?

Will be at least 67 mm? P =

(b) Suppose that a sample of 100 adult males is to be obtained. Without assuming that interpupillary distance is normally distributed, what is the approximate probability that the sample average distance will be between 64 and 66 mm? (Round all your intermediate calculations to four decimal places. Round the answers to four decimal places.) P =

Will be at least 67 mm? P =

Explanation / Answer

(a) So the probability that the sample average distance for these 25 will be between 63 and 66 mm is

P(63<xbar<66) = P((63-65)/(5/sqrt(25)) <(xbar-mean)/(s/vn) <(66-65)/(5/sqrt(25)))

=P(-2<Z<1) =0.8186 (from standard normal table)

Will be at least 67 mm?

P(X>67) = P(Z>(67-65)/(5/sqrt(25)))

=P(Z>2) =0.0228(from standard normal table)

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(b) So the probability that the sample average distance for these 25 will be between 63 and 66 mm is

P(63<xbar<66) = P((63-65)/(5/sqrt(25)) <(xbar-mean)/(s/vn) <(66-65)/(5/sqrt(25)))

=P(-2<Z<1) =0.8186 (from standard normal table)

Will be at least 67 mm?

P(X>67) = P(Z>(67-65)/(5/sqrt(25)))

=P(Z>2) =0.0228(from standard normal table)

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