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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?

andselisk
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  • the probabilities are equal implies the exponents are equal. Equate them and solve for the value at which they will be equal. – tpb261 Sep 08 '14 at 11:43
  • Which exponents are you talking about exactly? –  Sep 08 '14 at 11:47
  • It's normal distributions so the pdf would be an exponential fn. of the for $exp{-\frac{(x-\mu)}{\sigma^2}$, right? – tpb261 Sep 08 '14 at 11:50
  • We want $c$ to be just as many $A$ standard deviation units above $10$ as $c$ is below $14$ in $B$ standard deviation units. So the interval from $10$ to $14$ should be split in the ratio of $1$ to $2$. – André Nicolas Sep 08 '14 at 11:50

2 Answers2

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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.

callculus42
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  • Right, sorry, I should've have mentioned that this is the point that I actually got stuck at. From here, how can I solve for c? My calculator has two options: normCDf which finds the prob. between a range of values and inverseNormCDf, which is the opposite (for left and right). My calculator will not let me equate these two to solve. I must hand it a definite probability (e.g. 0.5, 0.765 etc., no variables). –  Sep 08 '14 at 13:20
  • @BirdinaForest You don´t really need to deal with NormCDF. $\Phi\left( \frac{c- \mu_a}{\sigma_a} \right)=\Phi\left( \frac{ \mu_b-c}{\sigma_b} \right)$ Taking on both sides the inverse of $\Phi(.)$ gives $\frac{c- \mu_a}{\sigma_a}=\frac{ \mu_b-c}{\sigma_b}$. This equation can be solved, more or less, easily. But I´m not sure, if I understood you correctly. – callculus42 Sep 08 '14 at 13:29
  • I think that clears everything up. I completely overlooked the fact that you could just inverse both sides. Thank you for the answer, I appreciate it. –  Sep 08 '14 at 14:37
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If you have access to EXCEL and can be used to solving @Calculus' equality, you can set it up like this and calculate the value of c

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Satish Ramanathan
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    I'm not interested in using Goal Seek to find a value. I want to know how to find the value mathematically. –  Sep 08 '14 at 13:16
  • I think the other responder has given you the solution which is $\frac{(c-10)}{1} = \frac{(14-c)}{2} => 2c-20 = 14-c => 3c = 34 => c = 34/3$ – Satish Ramanathan Sep 08 '14 at 13:43