Question

A state insurance commission estimates that 13% of all motorists in its state areuninsured. Suppose this proportion is valid. Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured. You may assume that the normaldistribution applies.

Answer 1

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

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 = 50, p = 0.13`

So

`\mu = E(X) = np = 50*0.13 = 6.5`

`\sigma = √(V(X)) = √(np(1-p)) = √(50*0.13*0.87) = 2.38`

Find the probability that in a randomsample of 50 motorists, at least 5 will be uninsured.

This is 1 subtracted by the pvalue of Z when X = 4. So

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

`Z = (4 - 6.5)/(2.38)`

`Z = -0.84`

`Z = -0.84` has a pvalue of 0.2005

1 - 0.2005 = 0.7995

79.95% probability that in a randomsample of 50 motorists, at least 5 will be uninsured.