Question

The heights of North American women are normally distributed with a mean of 64 inches and a standard deviation of 2inches.a. What is the probability that a randomly selected woman is taller than 66 inches?b. A random sample of four women is selected. What is the probability that the sample mean height is greater than 66inches?c. What is the probability that the mean height of a random sample of 100 women is greater than 66 inches?

Answer 1

Answer:

a)$P(X>66)=P(Z>1)=1-P(Z$

b)$P(\bar X >66)=P(Z>2)=1-P(Z$

c) $P(\bar X >66)=P(Z>10)=1-P(Z$

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".  

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".

2) Part a

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

$X \sim N(64,2)$  

Where $\mu=64$ and $\sigma=2$

We are interested on this probability

$P(X>66)$

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:

$P(X>66)=P(\frac{X-\mu}{\sigma}>\frac{66-\mu}{\sigma})=P(Z>\frac{66-64}{2})=P(Z>1)$

And we can find this probability on this way:

$P(Z>1)=1-P(Z$

3) Part b

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

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

$P(\bar X >66)=P(Z>\frac{66-64}{\frac{2}{\sqrt{4}}}=2)$

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

$P(Z>2)=1-P(Z$

4) Part c

$P(\bar X >66)=P(Z>\frac{66-64}{\frac{2}{\sqrt{100}}}=10)$

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

$P(Z>10)=1-P(Z$