Question

In Excel we will use the command/function NORM.INV to get the inverse cdf of normal distributions. This command takes a left tail area (between 0 and 1) as it's first argument. The weights of bags of cookies are normally distributed with a mean of 15 ounces and a standard deviation of 0.85 ounces. Bags of cookies that have weights in the upper 7.5% are too heavy and must be repackaged. What is the most a bag of cookies can weigh and not need to be repackaged?

Answer 1

The maximum bag of cookies weight allowed not to be repackaged is 16.223 ounces

Explanation:

`\mu=15\\\sigma= 0.85\\`

1. Probability equation

We need to find the maximum weight a bag of cookies allowed not to be repackaged.

Bags with weights in the upper 7.5% must be repackaged. It means that the percentage of bags must not be repackaged is:

`P(Not repackaged)=1-P(repackaged)\\P(Not repackaged)=1-0.075\\P(Not repackaged)=0.925`

This probability is equivalent to
`P(Z\leq 0.925)` in a standard normal distribution.

2. Get the inverse cdf

Excel function DISTR.NORM.INV estimate the inverse of the normal cumulative distribution for the mean and standard deviation.

In this case, it means the function will estimate the X- value corresponding to
`P(Z\leq 0.925)` from a normal standard distribution with
`\mu=15` and
`\sigma=8`.

Function DISTR.NORM.INV has the structure:

`=DISTR.NORM.INV(P;\mu;\sigma)`

When:

P: Probability

μ: Mean

σ: Standard Deviation

Then, we obtain this structure by replacing the dates:

`=DISTR.NORM.INV(0,925;15;0,85)\\=16,223`

The maximum bag of cookies weight allowed not to be repackaged is 16.223 ounces

Answer 2

`z=1.439<(a-15)/(0.85)`

And if we solve for a we got

`a=15 +1.439*0.85=16.223`

So the value of height that separates the bottom 92.5% of data from the top 7.5% is 16.223.

And using the NORM.INV function in excel we have:

"=NORM.INV(1-0.075,15,0.85)"

Snd we got the same result

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(15,0.85)`

Where
`\mu=15` and
`\sigma=0.85`

For this part we want to find a value a, such that we satisfy this condition:

`P(X>a)=0.075` (a)

`P(X<a)=0.925` (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 0.925 of the area on the left and 0.075 of the area on the right it's z=1.439. On this case P(Z<1.439)=0.925 and P(z>1.439)=0.075

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

`P(X<a)=P((X-\mu)/(\sigma)<(a-\mu)/(\sigma))=0.925`

`P(z<(a-\mu)/(\sigma))=0.925`

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

`z=1.439<(a-15)/(0.85)`

And if we solve for a we got

`a=15 +1.439*0.85=16.223`

So the value of height that separates the bottom 92.5% of data from the top 7.5% is 16.223.

And using the NORM.INV function in excel we have:

"=NORM.INV(1-0.075,15,0.85)"

Snd we got the same result