a Credit: CLT .1 The National Health and Nutrition Examination Survey of 1988-19

Question

a Credit: CLT .1 The National Health and Nutrition Examination Survey of 1988-1994 (NHANES III A-1) estimated the mean serum cholesterol level for U.S. females aged 20-74 years to be 204 mg/dl. The estimate of the standard deviation was approximately 44, Using these estimates as the mean ? and standard deviation ? for the U.S. population, consider the sampling distribution of the sample mean based on samples of size 50 drawn from women in this age group. What is the mean of the sampling distribution? The standard error? 3.2 The study cited in Exercise 5.3.1 reported an estimated mean serum cholesterol level of 193 for rs. The estimated standard deviation was approximately 37. Use these women aged 20-29 yea estimates as the mean ? and standard deviation ? for the US. population. If a simple random sample of size 60 is drawn from this population, find the probability that the sample mean serum cholesterol level will be: (a) Between 170 and 195 (c) Greater than 190 only (b) Below 175

Explanation / Answer

3.1
mean of sampling distribution = 204
std. error = 44/sqrt(50) = 6.2225

3.2
mean = 183
sd = 37
n = 60

a)
P(170 < X < 195)
= P(X < 195) - P(X < 170)
= P(z < (195 - 183)/(37/sqrt(60))) - P(z < (170 - 183)/(37/sqrt(60)))

= P(z < 2.5122) - P(z < -2.7216)
= 0.9940 - 0.0032
= 0.9908

b)
P(X < 175)
= P(z < (175 - 183)/(37/sqrt(60)))
= P(z < -1.6748)
= 0.047

c)
P(X > 190)
= P(z > (190 - 183)/(37/sqrt(60)))
= P(z > 1.4655)
= 0.0714

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