Factor 10n^2 - 37n-36
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I assume otherwise. If
is indeed a proper PDF, then its integral over the support of
is 1.
Compute the integral.
Then
The number of randomly selected employers that would have to be contacted is 9604
<h3>How to solve for the value of n</h3>The confidence level is 95%
1 - 0.95 = 0.05
We have to get the z value at 0.05
0.05/2 = 0.025
Using the standard normal table, the critical value at 0.025 = 1.96
We have to define the probability as P = 50% = 0.5
we have to define q = 1 - 50% = 0.5
The formula that would be used to find the value of n is given as
These variables used are:
pq = variance of population
E = margin of error = 0.01
critical value = Zc = 1.96
= 0.25 * 38416
= 9604
Hence the sample space is given as 9604
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