Question

The heights of baby giraffe are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches. If 100 baby giraffes are randomly selected, find the probability that they have a mean height less than 63.0 inches.

Answer 1

Answer:

$P(\bar X$

And we can solve this using the following z score formula:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we use this formula we got:

$z = \frac{63-63.6}{\frac{2.5}{\sqrt{100}}}= -2.4$

So we can find this probability equivalently like this:

$P( Z$

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".  

Solution to the problem

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:

$X \sim N(63.6,2.5)$  

Where $\mu=63.6$ and $\sigma=2.5$

We select n =100. Since the distribution for X is normal then we know that the distribution for the sample mean $\bar X$ is given by:

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

We want this probability:

$P(\bar X$

And we can solve this using the following z score formula:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we use this formula we got:

$z = \frac{63-63.6}{\frac{2.5}{\sqrt{100}}}= -2.4$

So we can find this probability equivalently like this:

$P( Z$