Answer:
(a) 0.5398
(b) 0.06376
Step-by-step explanation:
Given u=118, SD = 25, n = 10
Since we are dealing with a stratified random sampling (SRS)
We use,
Z = X - U/(Sd/√n).
The required probability =
Pr(112 < Z < 114)
Pr(112< Z< 114)=Pr( Z<114) - Pr(Z<112)
=Pr [Z<(114-118)/(25/√100)] - Pr[Z<(112-118)/(25/√100)]
= Pr(Z<-1.6) - PR(Z<-2.4)
= 0.5498 - 0.00820
= 0.5398
(b) Now since we are considering the whole population of 1000 men we use
Z = (X - U)/Sd
Now we the Probability of Pr(118±2)= Pr(120, 116)
= Pr(116 < X <120)
Pr(118±2) = Pr(Z<120-118/25) - Pr(Z<116-118/25)
Pr(118±2) = Pr(Z<0.08) - Pr(Z< -0.08)
= 0.53188 - 0.46812
Pr(118±2)= 0.06376
Please note the complete form of this question was gotten through Google.
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
x takes a value between 114 and 122 mg/dL? This is the probability that
¯¯¯
x
x estimates μ within ±4 mg/dL.
(b) Choose an SRS of 1000 men from this population. Now what is the probability that
¯¯¯
x
x falls within ±4 mg/dL of μ? The larger sample is much more likely to give an accurate estimate of μ.
