The amount of a certain chemical in a type $A$ cell is normally distributed with mean of $10$ and a standard deviation of $1$, while 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 a type $A$ or a type $B$, the amount of chemical in the cell is measured and the cell is classified as a type $A$ if the amount is less than a specified value $c$, and as being of type $B$ otherwise.
Find the value of $c$ for which the two probabilities of misclassification are equal.
May I please get help with this question? My working:
$$ \begin{align} A &\sim N(10, 1^2)\\ B &\sim N(14, 2^2)\\ P(A>c) &= P(B<c) \end{align} $$
Here's where I'm stuck. How am I supposed to know what the probability is so that I can use my calculator to solve for $c$? (Using the inverse normal distribution function). I am unable to simply solve this equation above, I must hand the calculator a number in the real set. That means, I cannot ask it to solve for a variable, but I can get it to find $c$ if and only if I know what the probability is by using the inverse normal distribution function. Equating these two does not give me a probability to work with and so I cannot use my calculator to solve it.
Microsoft Excel is not an option in a test situation, so that is irrelevant.
The textbook supplies $\frac{34}{3}$ as the answer. How did they get an exact answer like that?
For type A it is $P(X_a \leq c) =\Phi\left( \frac{c- \mu_a}{\sigma_a} \right)$
For type B it is $P(X_b \geq c) =1-\Phi\left( \frac{c- \mu_b}{\sigma_b} \right)=\Phi\left( \frac{ \mu_b-c}{\sigma_b} \right)$
$\Phi(.)$ is the cdf of the standard normal distribution.
This two expressions can be set equal and solved for c.