Question

A Snapshot in USA Today indicates that 51% of Americans say the average person is not very considerate of others when talking on a cellphone. Suppose that 100 Americans are randomly selected. Find the approximate probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone. (Use the normal approximation. Round your answer to four decimal places.)

Answer 1

There is an 1.39% probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone.

Explanation:

For each American, there are only two possible outcomes. Either they indicate that the average person is not very considerate of others when talking on a cellphone, of they indicate that the average person is considerate. This means that we use the binomial probability distribution to solve this problem.

This distribution 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:

There are 100 americans, so
`n = 100`

51% of Americans say the average person is not very considerate of others when talking on a cellphone. This means that
`p = 0.51`.

Find the approximate probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone.

This is 1 subtracted by the pvalue of Z when
`X = 62`.

We have that

`\mu = E(X) = np = 100*0.51 = 51`

`\sigma = √(V(X)) = √(np(1-p)) = √(100*0.51*0.49) = 5`

So

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

`Z = (62 - 51)/(5)`

`Z = 2.2`

`Z = 2.2` has a pvalue of 0.9861.

This means that there is a 1-0.9861 = 0.0139 = 1.39% probability that 62 or more Americans would indicate that the average person is not very considerate of others when talking on a cellphone.