Chapter 11.10, Problem 32PPSE

Solution

To calculate: The sample size for the normal distribution of the weight of tubs of yogurt.

The required sample size for the given scenario is 480.

Given:The distribution of the weight of tubs of yogurt is normal with a mean of 1.0 lb. and a standard deviation of 0.06 lb.

The number of tubs less than 0.88 lb. is 12.

Concept used:Z score is the number of standard deviations away from the mean and using a standard normal table probability of data less than a particular Z score can be calculated. The formula for Z score is

  $\begin{array}{l}Z=\frac{X-\mu }{\sigma }\\ \text{where X is the score for which Z score is calculated}\text{.}\\ \mu \text{ is the mean and }\sigma \text{ is the standard deviation}\text{.}\end{array}$

With reference to the empirical rule within two standard deviations of the mean the percentage of data is 95% and each tail has about 2.5% of the data.

Calculation:

The Z score for weight 0.88 lb. is calculated as shown below

  $\begin{array}{l}Z=\frac{X-\mu }{\sigma }\\ Z=\frac{0.88-1}{0.06}=\frac{-0.12}{0.06}=-2\end{array}$

The percentage of data less than a Z score of -2 is 2.5% with respect to the empirical rule. Thus the probability that a weight is less than 0.88 lb. or Z score -2 is 0.025.

There are total of 12 yogurt tubs less than 0.88 lb. which is 2.5% of total sample size n.

  $\begin{array}{l}0.025=\frac{12}{n}\\ n=\frac{12}{0.025}=480\end{array}$

2.5% of 480 is 12. Thus the total required sample size is 480.