Statistics Problem Need Help!

Question

Suppose the amount of a popular sport drink in bottles leaving the filling machine has a normal distribution with mean 101.5 ml and standard deviation of 1.6 ml.

 

A.  If the bottles are labeled 100 ml, what proportion of the bottles contain less than the labeled amount?

 

B.  If four bottles are randomly selected, find the mean and standard deviation of the average content

 

C. What is the probability that the average content of four bottles is less than 100 ml?

Answer 1

Great questions. Here are my answers:

Suppose the amount of a popular sport drink in bottles leaving the filling machine has a normal distribution with mean 101.5 ml and standard deviation of 1.6 ml. 

We'll use this to set up the problems. (A) is looking for the proportion of bottles that have less than 100ml. (B) is asking about what the distribution of the sample mean will look like if we're talking about a random sample of four bottles and (C) is going to use the information from (B) to calculate the proportion of samples of size n=4 with means below 100ml.

A. If the bottles are labeled 100 ml, what proportion of the bottles contain less than the labeled amount? 

We need to use the information from the problem to get the Z-score for 100ml. We get this by finding (100 - 101.5) / 1.6 = -0.9375.

Pr( X < 100) = Pr(Z < -0.9375) = 0.1743

I used R to calculate this probability by entering pnorm(-0.9375) into the command line. Alternatively we can use this Excel function to get the same result:

=NORM.DIST(-0.9375,0,1,TRUE)

Paste this into an Excel sheet cell and the answer will appear. We could also use

=NORM.DIST(100,101.5,1.6,TRUE)

To get the answer without finding the Z-score.

B. If four bottles are randomly selected, find the mean and standard deviation of the average content

The distribution of the sample mean will always be at least as normal as the population the sample came from. Since our population of bottles is already normally distributed the sample mean will be normal as well no matter how big the sample size is. We also know that the mean of the sample means is the same as the population mean, which in this case is 101.5. The standard deviation of the means may be found by dividing the population SD by the square root of the sample size, or 1.6/sqrt(4) = 0.8.

Xbar (for n=4) ~ N(μ=101.5, σ=0.8)

 

C. What is the probability that the average content of four bottles is less than 100 ml?

Using the same approach with Excel we put

=NORM.DIST(100,101.5,0.8,TRUE)

into a cell and get Pr(Xbar < 100) = 0.0304