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Fri 17 Apr, 2015 07:23 pm
An event planner does research and finds that approximately 2.75% of the people in the area where a large event is being held are pescatarian. Treat the 250 guests expected at the event as a simple random sample from the local population of about 150,000. Assume the sampling distribution is approximately Normal.
Suppose the event planner assumes that 4% of the guests will be pescatarian so he orders 10 pescatarian meals. What is the approximate probability that more than 4% of the guests are pescatarian and that he will not have enough pescatarian meals?
@cynthdlt,
If P=2.75%, then the expected value is 2.75% x 250 = 6.875.
The expected standard deviation = sqrt[ p (1-p) N ] = sqrt [ 2.75% x 97.25% x 250 ] = 2.585
10 = 6.875 + 2.585 Z
Z = 1.21
If you look up 1.21 on a
Z score calculator you get 11.3% chance of not having enough meals.
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