Question

A government sample survey plans to measure the LDL (bad) cholesterol level of an SRS of 100 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 = 118 milligrams per deciliter (mg/dL) and standard deviation σ = 25 mg/dL.
(a) What is the probability that x¯ takes a value between 114 and 122 mg/dL? This is the probability that x¯ estimates μ within ±4 mg/dL.
(b) Choose an SRS of 1000 men from this population. Now what is the probability that x¯ falls within ±2 mg/dL of ???? ? (Enter your answer rounded to four decimal places.) probability: .9128 The larger sample is much more likely to give an accurate estimate of ????.

Answer 1

Answer:

Part a: The probability that x takes a value between 114 and 122 mg/dL is 0.8904.

Part b: The probability that x falls within ±2 mg/dL of μ is 0.9886.

Step-by-step explanation:

Part a

$sd = \sigma / \sqrt{n} = 25 / \sqrt{100} = 25/10 = 2.5$

Here we have to find P(114 < x¯ < 122).

Now convert x¯  = 114 and 122 into z-score.

z = (x¯ - μ) / sd

z = (114 - 118) / 2.5 = -1.6

z = (122 - 118) / 2.5 = 1.6

That is now we have to find P(-1.6 < Z < 1.6)

P(-1.6 < Z < 1.6) = P(Z < 1.6) - P(Z < -1.6)

This probability we can find by using EXCEL.

syntax :

=NORMSDIST(z)

where z is test statistic value.

P(-1.6 < Z < 1.6) = 0.9452 - 0.0548 = 0.8904

The probability that x takes a value between 114 and 122 mg/dL is 0.8904.

Part b

Here sample size = 1000

$sd = \sigma / \sqrt{n} = 25 / \sqrt{1000} = 0.7906$

Now convert Xbar = 114 and 122 into z-score.

z = (x¯ - μ) / sd

z = (116 - 118) / 0.7906 =-2.53

z = (120 - 118) / 0.7906= 2.53

That is now we have to find P(-2.53 < Z < 2.53)

P(-2.53 < z < 2.53) = P(z< 2.53) - P(z < -2.53)

= 0.9943 - 0.0057 = 0.9886

The probability that x falls within ±2 mg/dL of μ is 0.9886.