Question

Pew Research reported in 2012 that the typical response rate to their surveys is only 9%. If for a particular survey 13,000 households are contacted, what is the probability that at least 1,400 will agree to respond?

Answer 1

Can’t help dude sorry about that

Answer 2

0% probability that at least 1,400 will agree to respond

Explanation:

We 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:

`√(V(X)) = √(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 = (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 = √(V(X))`.

In this problem, we have that:

`n = 13000, p = 0.09`

So

`\mu = E(X) = np = 13000*0.09 = 1170`

`\sigma = √(V(X)) = √(np(1-p)) = √(13000*0.09*0.91) = 32.63`

If for a particular survey 13,000 households are contacted, what is the probability that at least 1,400 will agree to respond?

This probability is 1 subtracted by the pvalue of Z when X = 1400. So

`Z = (X - \mu)/(\sigma)`

`Z = (1400 - 1170)/(32.63)`

`Z = 7.05`

`Z = 7.05` has a pvalue of 1.

1 - 1 = 0

0% probability that at least 1,400 will agree to respond