The owner of a new restaurant wants to have seating for more than 75 people. There is currently a
2 answers:
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Answer:
a. If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.
b. If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.
Step-by-step explanation:
This problem should be solved with a binomial distribution sample, but as the size of the sample is large, it can be approximated to a normal distribution.
The parameters for the normal distribution will be
We can calculate the z values for x1=0.47 and x2=0.51:
We can now calculate the probabilities:
If a sample of size 200 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 41.26%.
b) If the sample size change, the standard deviation of the normal distribution changes:
We can calculate the z values for x1=0.47 and x2=0.51:
We can now calculate the probabilities:
If a sample of size 100 is taken, the probability that the proportion of successes in the sample will be between 0.47 and 0.51 is 30.5%.