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

`P(\bar X <63)`

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

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And if we use this formula we got:

`z = (63-63.6)/((2.5)/(√(100)))= -2.4`

So we can find this probability equivalently like this:

`P( Z<-2.4) = 0.0082`

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, (\sigma)/(√(n)))`

We want this probability:

`P(\bar X <63)`

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

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And if we use this formula we got:

`z = (63-63.6)/((2.5)/(√(100)))= -2.4`

So we can find this probability equivalently like this:

`P( Z<-2.4) = 0.0082`