Women have pulse rates that are normally distributed with a mean of 74.9beats pe

Question

Women have pulse rates that are normally distributed with a mean of 74.9beats per minute and a standard deviation of 11.31.

A;A doctor sees exactly 20patients each day. Find the probability that 20 randomly selected women have a mean pulse rate between 60 and 90beats per minute.

B.

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8614g and a standard deviation of 0.0515 g. A sample of these candies came from a package containing 470candies, and the package label stated that the net weight is 401.3g. (If every package has 470candies, the mean weight of the candies must exceed

Start Fraction 401.3 Over 470 EndFraction401.3470equals=0.8539g for the net contents to weigh at least 401.3g.

If 1 candy is randomly selected, find the probability that it weighs more than 0.8539g.what the probability

Explanation / Answer

Since = 74.9 and = 11.31 we have:

P ( 60 < X < 90 ) = P ( 6074.9 < X < 9074.9 )

= P ( 6074.9 / (11.31/sqrt(20)) < X / (/sqrt(n)) < 9074.9 / (11.31/sqrt(20))

Since Z = (x) / /sqrt(n) , (6074.9) / (11.31/sqrt(20)) =5.89 and (9074.9) / (11.31/sqrt(20)) = 5.97 we have:

P ( 60 < X < 90 )=P ( 5.89 < Z < 5.97 )

Using the standard normal table to conclude that:

P ( 5.89 < Z < 5.97 ) = 0.8165

b).

Since = 0.8614 and = 0.0515 we have:

P ( X > 0.8539 ) = P ( X > 0.85390.8614 ) = P ( X/ > 0.85390.8614 / 0.0515)

Since Z = x/ and (0.85390.8614) / 0.0515 = 0.15 we have:

P ( X > 0.8539 ) = P ( Z > 0.15 )

Using the standard normal table to conclude that:

P (Z > 0.15) = 0.5596

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