Question

Assume that weights of men are normally distributed with a mean of 170 lb and a standard deviation of 30 lb (approximate values based on data from the National Health and Nutrition Examination Survey). What percentage of individual men have weights less than 185 lb? If samples of 36 men are randomly selected and the mean weight is computed for each sample, what percentage of the sample means are less than 185 lb?
a. 0.3156
b. 0.2611
c. 0.2312
d. 0.4602

Answer 1

Answer:

a) $P(X<185)=P(Z<0.5)=0.691$

b) $P(\bar X <185)=P(Z<3)=0.99865$

Step-by-step explanation:

1) 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".  

2) Part a

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

$X \sim N(170,30)$  

Where $\mu=170$ and $\sigma=30$

We are interested on this probability

$P(X<185)$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(X<185)=P(\frac{X-\mu}{\sigma}<\frac{185-\mu}{\sigma})=P(Z<\frac{185-170}{30})=P(Z<0.5)$

And we can find this probability on this way:

$P(Z<0.5)=0.691$

3) Part b

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

$P(\bar X <185)=P(Z<\frac{185-170}{\frac{30}{\sqrt{36}}}=3)$

And using a calculator, excel or the normal standard table we have that:

$P(Z<3)=0.99865$