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

μ = 119 milligrams per deciliter (mg/dL) and standard deviation σ = 25 mg/dL. Use Table A for the following questions, where necessary.

(a) Choose an SRS of 100 men from this population. What is the sampling distribution of x? (Use the units of mg/dL.)
Answer = N(119,2.5)

(b) What is the probability that x takes a value between 116 and 122 mg/dL? This is the probability that x estimates μ within ±3 mg/dL. (Round your answer to three decimal places.)

I only need the answer to part (b) I asked this question before and the answer is NOT 0.096

Answer 1

Solution :

Given that,

mean = = 119

standard deviation = = 25

n = 100

= = 119

= / n = 25 / 100 = 2.5

P(116 < < 122)  

= P[(116 - 119) /2.5 < ( - ) / < (122 - 119) / 2.5)]

= P(-1.20 < Z < 1.20)

= P(Z < 1.20) - P(Z < -1.20)

Using z table,  

= 0.885 - 0.115

= 0.770