Solution:
Given that;
The proportion of individuals insured by the All-Driver Automobile Insurance Company who received at least one traffic ticket during a five-year period is .15
That is,
[tex]\mu=0.15[/tex]If a random sample of 150 insured individuals is used to estimate the proportion having received at least one ticket, the formula to show the sampling proportion is
[tex]\begin{gathered} \sigma=\sqrt{\frac{\mu(1-\mu)}{n}} \\ \sigma=\sqrt{\frac{0.15(1-0.15)}{150}}=0.0292 \end{gathered}[/tex]Hence, the answer is 0.0292 (four decimal places)
It is approximately normal.
b)
[tex]\begin{gathered} \frac{0.03}{0.0292}=1.0290 \\ \frac{-0.03}{0.0292}=-1.0290 \end{gathered}[/tex]The probability the sample proportion will be
[tex]2P\left(0\leq z\leq1.0292\right)=0.6966[/tex]Hence, the answer is 0.6966 (four decimal places)
