Respuesta :
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.
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
