Question

Two bottling plants package a certain type of sports drink. Suppose the mean volume of all of this type of sports drinks is 20 fluid ounces. Bottling plant A bottles an average of 50,000 sports drinks per day. Bottling plant B bottles an average of 175,000 sports drinks per day. On a particular day, which bottling plant is less likely to record a mean volume of 21 fluid ounces for the day?

Answer 1

Given Information:

Mean volume = μ = 20 fl oz

Sampling size of plant A = n₁ = 50,000 drinks/day

Sampling size of plant B = n₂ = 175,000 drinks/day

Required Information:

Which bottling plant is less likely to record a mean volume of 21 fl oz ?

Answer:

Plant B is less likely to record a mean volume of 21 fl oz

Step-by-step explanation:

The standard deviation of the plant A is given by

σa = σ/√n₁

The standard deviation of the plant B is given by

σb = σ/√n₂

Where σ is standard deviation for the mean volume of 20 fl oz and it is fixed.

As you can notice in the above relation, the standard deviation of the plants depend upon the number of samples (n). As the number of sample increases, the standard deviation of samples decreases which means that the mean of the samples will be closer to the actual mean (that is 20 fl oz).

Since the plant B has more samples (n₂ = 175,000) then its standard deviation (σb) will be less and the mean will be closer to 20 fl oz therefore, it is less likely that it will record a mean of 21 fl oz.

Plant A is more likely to record a mean volume of 21 fl oz

Answer 2

Answer:

Bottling plant B (with 175,000 sports drinks per day), because the daily mean will be closer to 21 fluid ounces with more sports drinks in the sample.