Question

A market research firm conducts telephone surveys with a 42% historical response rate. What is the probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions

Answer 1

3.44% probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions

Explanation:

For each individual surveyed, there are only two possible outcomes. Either they will cooperate, or they will not. The probability of an individual cooperating is independent from other individuals. So the binomial probability distribution is used to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

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:

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal.

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:

`\mu = E(X) = np = 400*0.42 = 168`

`\sigma = √(V(X)) = √(np(1-p)) = √(400*0.42*0.58) = 9.87`

What is the probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions

This probability is 1 subtracted by the pvalue of Z when X = 150. So

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

`Z = (150 - 168)/(9.87)`

`Z = -1.82`

`Z = -1.82` has a pvalue of 0.0344.

So 3.44% probability that in a new sample of 400 telephone numbers, at least 150 individuals will cooperate and respond to the questions