Question

An insurance agent research suggested that first year drivers had roughly an 13% chance of being involved in an automobile accident while driving. The insurance agent provided insurance to 152 first year drivers last year. What is the probability (based on the statistics) that 24 or more of those drivers were involved in accidents?

Answer 1

Probability that 24 or more of those drivers were involved in accidents is 0.15625.

Explanation:

We are given that an insurance agent research suggested that first year drivers had roughly an 13% chance of being involved in an automobile accident while driving.

The insurance agent provided insurance to 152 first year drivers last year.

Let
`\hat p` = sample proportion of drivers who were involved in accidents

The z score probability distribution for sample proportion is given by;

Z =
`\frac{\hat p-p}{\sqrt(\hat p(1-\hat p))/(n) {} }` ~ N(0,1)

where, p = population proportion of first year drivers involved in an automobile accident while driving = 13%

n = sample of first year drivers = 152

Now, probability that 24 or more of those drivers were involved in accidents is given by = P(
`\hat p`
`\geq`
`(24)/(152)`)

P(
`\hat p`
`\geq` 0.16) = P(
`\frac{\hat p-p}{\sqrt(\hat p(1-\hat p))/(n) {} }`
`\geq`
`\frac{0.16-0.13}{\sqrt(0.16(1-0.16))/(152) {} }` ) = P(Z
`\geq` 1.01) = 1 - P(Z < 1.01)

= 1 - 0.84375 = 0.15625

The above probability is calculated by looking at the value of x = 1.01 in the z table which has an area of 0.84375.

Hence, the required probability is 0.15625.