To calculate: The
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
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
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.
2.5% of 480 is 12. Thus the total required sample size is 480.
Chapter 11 Solutions
High School Math 2015 Common Core Algebra 2 Student Edition Grades 10/11
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