Question

A market research firm did a report about the residential telephone numbers in New York State, which says 29% of all residential telephone number in NY are unlisted. A telemarketing company uses random digit dialing equipment that dials the residential numbers at random, regardless of whether they are listed. The firm calls 2000 numbers in NY. What is the probability that at most 590 numbers called are unlisted

Answer 1

69.85% probability that at most 590 numbers called are unlisted

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:

`√(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:

`p = 0.29, n = 2000`

So

`\mu = E(X) = np = 2000*0.29 = 580[tex]</p><p>[tex]\sigma = √(V(X)) = √(np(1-p)) = √(2000*0.29*0.71) = 20.29`

What is the probability that at most 590 numbers called are unlisted

Using continuity correction, this is
`P(X \leq 590 + 0.5) = P(X \leq 590.5)`, which is the pvalue of Z when X = 590.5. So

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

`Z = (590.5 - 580)/(20.29)`

`Z = 0.52`

`Z = 0.52` has a pvalue of 0.6985

69.85% probability that at most 590 numbers called are unlisted