The quotient of 5n^2(m^2+2m-3)/10n(m^2-3m+2) divided by n^2(m^2-3m-18)/4n^2(m^2-8m+12)
1 answer:
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Answer:
0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%
Step-by-step explanation:
I am going to use the binomial approximation to the normal to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
Can be approximated to a normal distribution, using the expected value and the standard deviation.
The expected value of the binomial distribution is:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that ,
.
In this problem, we have that:
. So
If the manager is correct, what is the probability that the proportion of flops in a sample of 404 released films would be greater than 4%?
This is the pvlaue of Z when X = 0.04*404 = 16.16. So
has a pvalue of 0.0091
1 - 0.0091 = 0.9909
0.9909 = 99.09% probability that the proportion of flops in a sample of 404 released films would be greater than 4%