Question

A recent survey by the American Automobile Association revealed that 80% of teenage girls text while driving. You have been hired to do a safety presentation to a high school class of 100 teenage girls. Using technology (e.g. STATDISK), what is the probability that at least 85 of these girls text while driving

Answer 1

13.14% probability that at least 85 of these girls text while driving

Explanation:

I am going to use the normal approximation to the binomial 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 = 0.8, n = 100`

So

`\mu = E(X) = np = 100*0.8 = 80`

`√(V(X)) = √(np(1-p)) = √(100*0.8*0.2) = 4`

What is the probability that at least 85 of these girls text while driving

Using continuity correction, this is
`P(X \geq 85 - 0.5) = P(X \geq 84.5)`, which is the pvalue of Z when X = 84.5. So

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

`Z = (84.5 - 80)/(4)`

`Z = 1.12`

`Z = 1.12` has a pvalue of 0.8686

1 - 0.8686 = 0.1314

13.14% probability that at least 85 of these girls text while driving