T-Mobile LTE * 69% 10\' + 4:32 PM webassign.net Suppose the heights of 18-year-o
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T-Mobile LTE * 69% 10' + 4:32 PM webassign.net Suppose the heights of 18-year-old men are approximately normally distributed, with mean 68 inches and standard deviation 4 inches. (a) What is the probability that an 18-year-old man selected at random is between 67 and 69 inches tall? (Round your answer to four decimal places.) (b) If a andom sample of twelve 18-year-old men is selected, what is the probability that the mean height x is between 67 and 69 inches? (Round your answer to four decimal places.) (c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this? The probability in part (b) is much higher because the standard deviation is smaller for the x distribution. O The probability in part (b) is much higher because the mean is larger for the x distribution. O The probability in part (b) is much lower because the standard deviation is smaller for the x distribution. The probability in part (b) is much higher because the mean is smaller for the x distribution. O The probability in part (b) is much higher because the standard deviation is larger for the x distribution. Let x be a random variable that represents the level of glucose in the blood (milligrams per deciliter of blood) after a 12 hour fast. Assume that for people under 50 years old, x has a distribution that is approximately normal, with mean ·77 and estimated standard deviation ·39. A test result x 40 is an indication of severe excess insulin, and medication is usually prescribed (a) What is the probability that, on a single test, xExplanation / Answer
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0, standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 68, standard Deviation ( sd )= 4
a.
BETWEEN THEM, To find P(a < = Z < = b) = F(b) - F(a)
P(X < 67) = (67-68)/4
= -1/4 = -0.25
= P ( Z <-0.25) From Standard Normal Table
= 0.4013
P(X < 69) = (69-68)/4 = 1/4 = 0.25
= P ( Z <0.25) From Standard Normal Table
= 0.5987
P(67 < X < 69) = 0.5987-0.4013 = 0.1974
b.
mean of the sampling distribution ( x ) = 68
standard Deviation ( sd )= 4/ Sqrt ( 12 ) =1.1547
sample size (n) = 12
P(X < 67) = (67-68)/4/ Sqrt ( 12 )
= -1/1.1547 = -0.86603
= P ( Z <-0.86603) From Standard Normal Table
= 0.19324
P(X < 69) = (69-68)/4/ Sqrt ( 12 )
= 1/1.1547 = 0.86603
= P ( Z <0.86603) From Standard Normal Table
= 0.80676
P(67 < X < 69) = 0.80676-0.19324 = 0.61352
c.
the probability an part (b) is much higher because the standard
deviation is smaller for the x distnbutlon.
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