A government sample survey plans to measure the blood cholesterol level of an SR

Question

A government sample survey plans to measure the blood cholesterol level of an SRS of men aged 20 to 34.
Suppose that in fact the blood cholesterol level of all men aged 20 to 34 follows the Normal distribution with mean ? = 186 milligrams per deciliter (mg/dl) and standard deviation ? = 41 mg/dl. Use Table A for the following questions, where necessary.

Step 1:
Choose an SRS of 100 men from this population.
What is the sampling distribution of x? (Use the units of mg/dl.)

Step 2:
What is the probability that x takes a value between 183 and 189 mg/dl?
This is the probability that x estimates ? within ±3 mg/dl.

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.

The N(186, 4.1) distribution. The N(186, 0.41) distribution. The N(1.86, 0.41) distribution. The N(186, 41) distribution.

Explanation / Answer

Mean ( u ) =186
Standard Deviation ( sd )= 41/ Sqrt ( 100 ) = 4.1
a)
The N(186, 4.1) distribution

b)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 183) = (183-186)/41/ Sqrt ( 100 )
= -3/4.1
= -0.7317
= P ( Z <-0.7317) From Standard Normal Table
= 0.23217
P(X < 189) = (189-186)/41/ Sqrt ( 100 )
= 3/4.1 = 0.7317
= P ( Z <0.7317) From Standard Normal Table
= 0.76783
P(183 < X < 189) = 0.76783-0.23217 = 0.5346

c)
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 183) = (183-186)/41/ Sqrt ( 1000 )
= -3/1.2965
= -2.3139
= P ( Z <-2.3139) From Standard Normal Table
= 0.01034
P(X < 189) = (189-186)/41/ Sqrt ( 1000 )
= 3/1.2965 = 2.3139
= P ( Z <2.3139) From Standard Normal Table
= 0.98966
P(183 < X < 189) = 0.98966-0.01034 = 0.9793  

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