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 ±4 mg/dL of μ? The larger sample is much more likely to give an accurate estimate of μ.

Answer 1

Answer:

a) $P(114 \leq\bar X \leq 122) = P(frac{114-118}{2.5} \leq \bar X \leq \frac{122-118}{2.5})= P(-1.6 \leq Z \leq 1.6)$

And we can find this probability using the z score table or excel and we got:

$P(-1.6 \leq Z \leq 1.6)=P(Z<1.6) -P(Z<-1.6)= 0.945-0.0548=0.8904$

b) $P(114 \leq\bar X \leq 122) = P(frac{114-118}{0.7906} \leq \bar X \leq \frac{122-118}{0.7906})= P(-5.06 \leq Z \leq 5.06)$

And we can find this probability using the z score table or excel and we got:

$P(-5.06 \leq Z \leq 5.06)=P(Z<5.06) -P(Z<-5.06)\approx 1$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Part a

Let X the random variable that represent the cholesterol level of a population, and for this case we know the distribution for X is given by:

$X \sim N(118,25)$  

Where $\mu=118$ and $\sigma=25$

Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=118, \frac{\sigma}{\sqrt{n}}=\frac{25}{\sqrt{100}}=2.5)$

And the z score for this case would be given by:

$z = \frac{\bar X -\mu}{\sigma_{\bar X}}$

We want to find this probability:

$P(114 \leq\bar X \leq 122) = P(frac{114-118}{2.5} \leq \bar X \leq \frac{122-118}{2.5})= P(-1.6 \leq Z \leq 1.6)$

And we can find this probability using the z score table or excel and we got:

$P(-1.6 \leq Z \leq 1.6)=P(Z<1.6) -P(Z<-1.6)= 0.945-0.0548=0.8904$

Part b

For this case the sample size is n =1000 so then the new deviation for the sample mean would be: $\sigma_{\bar X} =\frac{25}{\sqrt{1000}}=0.7906$

And we want to find this probability:

$P(114 \leq\bar X \leq 122) = P(frac{114-118}{0.7906} \leq \bar X \leq \frac{122-118}{0.7906})= P(-5.06 \leq Z \leq 5.06)$

And we can find this probability using the z score table or excel and we got:

$P(-5.06 \leq Z \leq 5.06)=P(Z<5.06) -P(Z<-5.06)\approx 1$