\begin{tabular}{l} 9:37 \ 3.11 Auto insurance premiums: Suppose a \ newspaper article states that the distribution of \ auto insurance premiums for residents of California \ is approximately normal with a mean of $$ \$ 1650 $$. The \ article also states that $$ 25 \% $$ of California residents \ pay more than $$ \$ 1900 $$. \ d) Identify the standard deviation of insurance \ premiums in California. (Keep 1 decimal place) \ \hline 409.8 \ Submit All Parts \ 4 \end{tabular}

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UpStudy AI · Accepted Answer

**Step 1:** Let $$ X $$ be the cost of auto insurance premiums. We are given that  
$$
X \sim N(\mu, \sigma^2),
$$
with $$\mu = 1650$$.

**Step 2:** The problem states that $$ 25\% $$ of California residents pay more than $$\$1900$$. This means
$$
P(X > 1900) = 0.25.
$$
Thus,
$$
P(X \le 1900) = 1 - 0.25 = 0.75.
$$

**Step 3:** Convert to a standard normal variable using
$$
Z = \frac{X - \mu}{\sigma}.
$$
The probability statement becomes:
$$
P\left( Z \le \frac{1900 - 1650}{\sigma} \right) = 0.75.
$$

**Step 4:** Look up the $$z$$-score corresponding to a cumulative probability of 0.75. The $$z$$-score for which $$ P(Z \le z_{0.75}) = 0.75 $$ is approximately
$$
z_{0.75} \approx 0.6745.
$$

**Step 5:** Set up the equation:
$$
\frac{1900 - 1650}{\sigma} = 0.6745.
$$
Simplify the numerator:
$$
\frac{250}{\sigma} = 0.6745.
$$

**Step 6:** Solve for $$\sigma$$:
$$
\sigma = \frac{250}{0.6745} \approx 370.4.
$$

**Final Answer:** The standard deviation of insurance premiums in California is approximately $$\$370.4$$.
The standard deviation of insurance premiums in California is approximately \$370.4.

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