An insurance company checks police records on 582 accidents selected at random and notes that teenagers were at the wheel 15.6% of the time. The standard deviation is 1.2%. Determine the confidence interval that is likely to contain the percent of all teenagers at the wheel when an accident occurred.

At 95% confidence level, the confidence interval that is likely to contain the percent, , of all teenagers at the wheel when an accident occurred is given as follows;

12.7% < < 18.6%

Step-by-step explanation:

The parameters given are as follows;

Sample size, n = 582

Proportion of accidents where teenagers are at the wheel, = 15.6%

x = 0.156*582 ≈ 91

The formula for the confidence interval is as follows;

Where:

z = z value at the confidence level

We therefore have;

At a confidence level of 95%, z = 1.96

The confidence interval at 95% confidence level is therefore;

12.7% < < 18.6%.

what is the original number? you forgot to add it.