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Here is the exercise question I'm trying to answer:

A certain fast food outlet produces hamburgers with weights that are normally distributed with mean $\mu = 990g$ and standard deviation $\sigma = 20g$. Find $w$ such that $80\%$ of hamburgers produced will weigh at least $w$.

My initial thought was to use the equation $z = \frac{x-\mu}{\sigma}$, find where $z$ covers the bottom $20\%$ of the normal distribution, and then just solve for $x$. But I don't know how to find that $z$ value. Any help is appreciated.

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  • $\begingroup$ Most statistical packages have a pre-packaged inverse to the normal. In Excel, for instance, you want "=NORMINV(0.2,990,20)". But if all you have is the normal, you can still do it by hand numerically (linear interpolation from a table, for example). $\endgroup$
    – lulu
    Mar 14, 2016 at 15:22
  • $\begingroup$ The $z$-value is about $-0.84$. How you get it depends on whether you are using software or tables. And if you are using tables, it depends on the layout of the table. From most tables you can see that $80\%$ of the area under the normal is to the left of $0.84$. So by symmetry $80\%$ of the area is above $-0.84$, and therefore $20\%$ of the area is below $-0.84$. Note that the $-0.84$ is approximate, That's a big hamburger. $\endgroup$
    – André Nicolas
    Mar 14, 2016 at 15:22
  • $\begingroup$ Should the question not be: find $w$ such that $P(weight\geq w)=0.8$. Either that or assume that they have had time to produce a vary large amount of hamburgers $\endgroup$
    – Connor Bishop
    Mar 14, 2016 at 15:27

1 Answer 1

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The Comment by @AndreNicholas describes how to use a printed table to get the answer. This is an 'inverse' problem. You look in the body of the table for 80% and then find the z-value from the margins of the table.

As mentioned in the Comment by @Lulu, many software packages provide procedures for 'inverse CDF' or 'quantile' probability functions. When you use printed normal CDF tables, those tables are for the 'standard' normal distribution, so you need to standardize with $z = \frac{w - \mu}{\sigma}$ as you have suggested.

However, when you use software, many problems like yours can be solved without standardization. I illustrate software use with R and Minitab statistical software. Because you likely have access to Excel, maybe you can figure out how to solve your problem with that software.

Using Minitab: Use the command 'INVCDF' with the subcommand 'NORM' as follows:

 MTB > invcdf .80;
 SUBC> norm 990 20.

 Inverse Cumulative Distribution Function 

 Normal with mean = 990 and standard deviation = 20

 P( X <= x )        x
         0.8  1006.83

You can also use the menu path CALC > Probability > Normal and then fill in appropriate blank fields in the dialog box.

Using R: The inverse cdf (quantile) function is called qnorm, and you can specify values of $\mu$ and $\sigma$ as 'parameters' of that function:

 qnorm(.80, 990, 20)
 ## 1006.832

You could also use qnorm without parameters for mean and SD. Then R assumes standard normal:

 qnorm(.80)
 ## 0.8416212

Solve $z = .8416212 = \frac{w - 990}{20}$ for $w = 1006.832.$

Note: Using printed normal tables, you will likely not find $exactly$ .8000 in the body of the table. In the one I'm looking at, .7995 is closest, and it corresponds to $z = 0.84.$ Then $w = 1006.8,$ which should be close enough for practical purposes. You could use linear interpolation to get another decimal place. (BTW: The phrase "at least" in the question that prompted @ConnorBishop's Comment, might mean to pick the next larger number in the table, which is .8023, yielding $z = 0.85$ and $w = 1007.0.$)

Below are figures showing the PDFs for $Norm(0, 1)$ and $Norm(990,20).$ In each case, the area (i.e., probability) beneath the curve and to the left of the dotted vertical red line is .80.

enter image description here

The figures below show the CDFs for $Norm(0, 1)$ and $Norm(990,20).$ At left, the vertical dashed purple line is at $z = .8416$; at the right $w = 1006.83.$

enter image description here

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