Question

3.30 Survey response rate. Pew Research reported in 2012 that the typical response rate to their surveys is only 9%. If for a particular survey 15,000 households are contacted, what is the probability that at least 1,500 will agree to respond

Answer 1

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 15000, p = 0.09$

So

$\mu = E(X) = np = 15000*0.09 = 1350$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{15000*0.09*0.91} = 35.05$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1499 - 1350}{35.05}$

$Z = 4.25$

$Z = 4.25$ has a pvalue of 1.

1 - 1 = 0

0% probability that at least 1,500 will agree to respond