Answer:
0% probability that at least 1,500 will agree to respond
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:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
![n = 15000, p = 0.09](https://tex.z-dn.net/?f=n%20%3D%2015000%2C%20p%20%3D%200.09)
So
![\mu = E(X) = np = 15000*0.09 = 1350](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%2015000%2A0.09%20%3D%201350)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15000*0.09*0.91} = 35.05](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B15000%2A0.09%2A0.91%7D%20%3D%2035.05)
What is the probability that at least 1,500 will agree to respond
This is 1 subtracted by the pvalue of Z when X = 1500-1 = 1499. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{1499 - 1350}{35.05}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1499%20-%201350%7D%7B35.05%7D)
![Z = 4.25](https://tex.z-dn.net/?f=Z%20%3D%204.25)
has a pvalue of 1.
1 - 1 = 0
0% probability that at least 1,500 will agree to respond