Question

A particular fruit's weights are normally distributed, with a mean of 239 grams and a standard deviation of 23 grams. If you pick 25 fruits at random, then 10% of the time, their mean weight will be greater than how many grams

Answer 1

`z=1.28<(a-239)/(4.6)`

And if we solve for a we got

`a=239 +1.28*4.6=244.89`

So the value of height that separates the bottom 90% of data from the top 10% is 244.89.

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(239,23)`

Where
`\mu=239` and
`\sigma=23`

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)))`

And we are interested on a value a such that:

`P(\bar X>a)=0.10` (a)

`P(\bar X<a)=0.90` (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

`P(\bar X<a)=P((\bar X-\mu)/((\sigma)/(√(n)))<(a-\mu)/((\sigma)/(√(n))))=0.9`

`P(z<(a-\mu)/((\sigma)/(√(n))))=0.9`

But we know which value of z satisfy the previous equation so then we can do this:

`z=1.28<(a-239)/(4.6)`

And if we solve for a we got

`a=239 +1.28*4.6=244.89`

So the value of height that separates the bottom 90% of data from the top 10% is 244.89.