A government sample survey plans to measure the LDL (bad) cholesterol level of a
ID: 3216730 • Letter: A
Question
A government sample survey plans to measure the LDL (bad) cholesterol level of an SRS of men aged 20 to 34. Suppose that in fact the LDL cholesterol level of all men aged 20 to 34 follows the Normal distribution with mean = 115 milligrams per deciliter (mg/dL) and standard deviation = 25 mg/dL. Use Table A for the following questions, where necessary. Please reffer to the internet for table A
Step 1: Choose an SRS of 100 men from this population. What is the sampling distribution of x? (Use the units of mg/dL.) The N(115, 2.5) distribution. The N(115, 0.25) distribution. The N(1.15, 0.25) distribution. The N(115, 25) distribution.
Step 2: What is the probability that x takes a value between 112 and 118 mg/dL? This is the probability that x estimates within ±3 mg/dL. 0.4654 0.7698 0.2327 0.6800
Step 3: Choose an SRS of 1000 men from this population. Now what is the probability that x falls within ±3 mg/dL of ? Use Table A and give your answer to 3 decimal places.
Explanation / Answer
= 115
= 25
1) n = 100
x = = 115
x = / sqrt(n) = 25 / sqrt(100) = 2.5
So, Option-A) The N(115, 2.5) distribution
2) P( within ±3 ) = P(-3/2.5 < Z < 3/2.5)
= P(-1.2 < Z < 1.2)
= P(Z < 1.2) - P(Z < -1.2)
= 0.8849 - 0.1151
= 0.7698
Option-B) 0.7698
3) n = 1000
x = / sqrt(n) = 25 / sqrt(1000) = 0.79
P( within ±3 ) = P(-3/0.79 < Z < 3/0.79)
= P(-3.8 < Z < 3.8)
= P(Z < 3.8) - P(Z < -3.8)
= 1 - 0
= 1 (approximaely)