Question

Suppose that in fact, the blood cholesterol level of all men aged 20-34 follows the Normal distribution with mean µ = 186 milligrams per deciliter (mg/dl) and standard deviation ? = 41 mg/dl.

a) Choose an SRS of 100 men from this population. What is the sampling distribution of x?? What is the probability that x takes a value between 183 and 189 mg/dl? This is the probability that x? estimates µ within ± 3 mg/dl.

b) Choose an SRS of 1000 men from this population. Now, what is the probability that x? falls within ± 3 mg/dl of µ? The larger sample is much more likely to give an accurate estimate of µ.

Answer 1

a) = = 186

   = = 41/sqrt(100) = 4.1

P(183 < < 189)

= P((183 - )/() < ( - )/( ) < (189 - )/( ))

= P((183 - 186)/4.1 < Z < (189 - 186)/4.1)

= P(-0.73 < Z < 0.73)

= P(Z < 0.73) - P(Z < -0.73)

= 0.7673 - 0.2327

= 0.5346

b) P(183 < < 189)

= P((183 - )/() < ( - )/( ) < (189 - )/( ))

= P((183 - 186)/(41/sqrt(1000)) < Z < (189 - 186)/(41/sqrt(1000)))

= P(-2.31 < Z < 2.31)

= P(Z < 2.31) - P(Z < -2.31)

= 0.9896 - 0.0104

= 0.9792