Question: The amount of a certain chemical in a type A cell is normally distributed with a mean of $10$ and a standard deviation of $1$. The amount in a type B cell is normally distributed with a mean of $14$ and a standard deviation of $2$. To determine whether a cell is type A or type B, the amount of chemical in the cell is measured. The cell is classified as type A if the amount is less than a specified value $c$, and as type B otherwise. Find the value of $c$ for which the two probabilities of misclassification are equal.

1. **State the problem:** We have two normal distributions: - Type A: $X_A \sim N(10, 1^2)$ - Type B: $X_B \sim N(14, 2^2)$ We classify a cell as type A if the chemical amount $< c$, and type B if $\geq c$. We want to find $c$ such that the probability of misclassifying a type A cell as type B equals the probability of misclassifying a type B cell as type A. 2. **Define misclassification probabilities:** - Misclassifying type A as B: $P(X_A \geq c) = 1 - P(X_A < c) = 1 - \Phi\left(\frac{c - 10}{1}\right)$ - Misclassifying type B as A: $P(X_B < c) = \Phi\left(\frac{c - 14}{2}\right)$ Here, $\Phi$ is the cumulative distribution function (CDF) of the standard normal distribution. 3. **Set the misclassification probabilities equal:** $$ 1 - \Phi\left(\frac{c - 10}{1}\right) = \Phi\left(\frac{c - 14}{2}\right) $$ 4. **Use the property of the standard normal CDF:** Since $1 - \Phi(z) = \Phi(-z)$, rewrite the equation as: $$ \Phi\left(-\frac{c - 10}{1}\right) = \Phi\left(\frac{c - 14}{2}\right) $$ 5. **Because $\Phi$ is strictly increasing, equate the arguments:** $$ -\frac{c - 10}{1} = \frac{c - 14}{2} $$ 6. **Solve for $c$:** Multiply both sides by 2: $$ -2(c - 10) = c - 14 $$ Simplify: $$ -2c + 20 = c - 14 $$ Bring all terms to one side: $$ -2c - c = -14 - 20 $$ $$ -3c = -34 $$ Divide both sides by $-3$: $$ c = \frac{-34}{-3} = \frac{34}{3} \approx 11.333 $$ 7. **Interpretation:** The threshold $c$ that equalizes the misclassification probabilities is approximately $11.333$. **Final answer:** $$ c = \frac{34}{3} \approx 11.333 $$