Statistic Question

Question

A machine fills bags of candy. Due to slight irregularities in the operation of the machine, not every bag gets exactly the same number of pieces. Assume that the number of pieces per bag is normally distributed, and has a mean of 378 with a standard deviation of 5.  a. What is the percentage of bags that contain between 378 and 393 pieces of candy? b. What is the percentage of bags have more than 373 pieces of candy? c. If 550 bags were sampled, how many bags would you expect to have less than 385 pieces of candy? 

Answer 1

Hello Azee,

We are given that the machine fills candy bags with a random number of pieces of candy that follows a normal distribution with mean 378 and standard deviation 5. We want to know the percentage of bags that will have between 378 and 393 pieces of candy.

To visualize the problem, sketch a bell curve with the highest point of the curve (the mode) over 378 on the horizontal axis. The total area under the bell curve is equal to 1.0, and we want to know the area that is between 378 and 393, or the area in the interval [378, 393].

We will use Excel to do this, but there may come a time when you have to use a table of the standard normal distribution, such as if you are taking an exam and do not have access to Excel, so I will show to do this first. Back in the day, when we didn't have Excel, this was the only way to do this (and we liked it :) ).

Anyway, the distribution we are given is a normal distribution with mean 378 and standard deviation 5 and we want to convert this to a standard normal random variable, where a standard normal random variable has a mean of zero and a standard deviation of 1. To convert between different normal distributions to a standard normal distribution, take the variable you have and subtract the mean of the distribution you are given and divide this quantity by the standard deviation, as follows:

z = standard normal random variable = (original variable – mean)/standard deviation.

The variable z will then follow a normal distribution with mean 0 and standard deviation 1, the so –called standard normal distribution. By the way, this computation is referred to as standardizing the variable.

In this problem, for the left point of the interval 378, the computation is

Z1 = (378-378)/5 = 0

For the right endpoint, 393, the computation is

Z2 = (393-378)/5 = 3.

The area under the standard normal curve between 0 and 3 is the same as the area between 378 and 393 for a normal distribution with mean 378 and standard deviation 5. Both of these areas are the area from the mean to 3 standard deviations to the right of the mean.

Tables of the standard normal distribution can be set up different ways, but usually provide the area under the standard normal curve to the left of a point on the horizontal axis, with that point usually being listed in the first column of the table. In the table then, you would go to the 3.0 in the first column and look inside the table (at 3.00, where the second decimal place 0 is often obtained from columns that run from 0 - 9 across the top of the table) to find that all of the area under the curve for z less than or equal to 3.0 is given by 0.9987. But since we only want the area between 3.0 and 0.0, we next need to subtract from that all of the area to the left of 0.0, which is of course 0.5. That leaves us with

Area between 378 and 393 for normal variable with mean 378 and standard deviation 5 =
Area between 0 and 3 for normal variable with mean 0 and standard deviation 1 = .9987 – 0.5 = 0.4987.

So this means that the answer  for (a) is that 49.87% of the candy bags will have between 378 and 393 pieces.

This is of course much easier to do in Excel. In Excel, the "NORM.DIST" function indicates the area to the left of a value for any normal distribution, and it has the arguments (i) value that you want to compute the area under the curve for, (ii) mean of standard normal distribution, (iii) standard deviation of the normal distribution and (iv) TRUE if you want the area under the curve or FALSE if you want the height of the normal distribution at that point. So to obtain the 0.9987 you can paste "=NORM.DIST(3,0,1,TRUE)" into the formula bar, and to get the 0.5, paste " =NORM.DIST(0,0,1,TRUE)" in, and then subtract the 0.5 from the 0.9987. In Excel, you will also get this answer if you paste in "=NORM.DIST(393, 378,5,TRUE)" to obtain the 0.9987 and then " =NORM.DIST(378, 378,5,TRUE)". So if you use Excel you do not even have to standardize the variable, which you would have to do if you use a table. However, it is worthwhile to know how to standardize a variable for other problems you will see (and in case you do not have Excel handy but are carrying a table of the standard normal distribution).

 

I will just work through the Excel for the rest of the parts.

 

b) Prob(X>373) = 1 - Prob( X less than or equal to 373) =1 - NORMDIST(373,378,5,TRUE) = 0.8413

 

or 84.13% of bags will have 373 or more pieces. Again, the NORMDIST function gives you the area under the curve that is less than or equal to 373 and you want the area to the right of this, which is why we subtract from 1.0

 

c. The probability a bag will have385 pieces or less is given by

 

Prob(X less than or equal to 385) = =NORMDIST(385,378,5, TRUE) = 0.9192,

 

so 91.92%t of the bags will have 385 pieces or fewer. Now 550 x 0.9192 = 505.58, so round to 506 bags that we would expect to have 385 pieces or fewer.

Hope this helps, but let me know if you have any questions.

Kind regards,
Hugh