Question

An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 130 lb and 181 lb. The new population of pilots has normally distributed weights with a mean of 139 lb and a standard deviation of 28.1 lb.A) If a pilot is randomly selected, find the probability that his weight is between 140 and 181 lb.B) If 32 different pilots are randomly selected, find the probability that their mean weight is between 140 and 181 lb

Answer 1

Answer:

a) $P(140$

$P(0.036$

$P(0.036$

b) $P(140< \bar X$

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 weights of a population, and for this case we know the distribution for X is given by:

$X \sim N(139,28.1)$  

Where $\mu=139$ and $\sigma=28.1$

We are interested on this probability

$P(140$

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:

$P(140$

And we can find this probability like this:

$P(0.036$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(0.036$

Part b

For this case we select a sample size of n =32. Since the distribution for X is normal then the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=139, \frac{\sigma}{\sqrt{n}}=\frac{28.1}{\sqrt{32}}=4.97)$

And the new z score would be:

$Z=\frac{\bar X -\mu}{\sigma_{\bar x}}$

$P(140< \bar X$

$P(140< \bar X$