Calculating the PPV and NPV using Bayes' Theorem A kit manufacturer markets a diagnostic kit for use in mass screening for
Thyroid disease. The diagnostic kit has the following specification: 
Diseased Population: Has response variable of $N(11,64)$ 
Healthy Population:  Has response variable of $N(3,64)$ 
Where the response variable is the measured thyroid level in the patients’ sample. 
$Q1$.) Show that the positive predictive value is $0.0437$ and the negative predictive value is $0.991$ if the prevalence of the disease is $2\%$
Applying Bayes' Theorem: $$\text{PPV} = \frac{0.02 \times \text{SE}}{0.02 \times \text{SE} + (1-0.02)(1-\text{SP})} $$
$$\text{NPV} = \frac{0.98 \times \text{SP}}{0.98 \times \text{SE} + (0.02)(1-\text{SE})} $$
I've tried doing simultaneous equations using MATLAB which gives $SE = 0.692651$ and $SP = 0.690663$ but I'm not sure how to calculate these values? 
 A: There are three issues left unresolved, one minor and the others somewhat more difficult.
The minor difficulty is whether the second parameter of the normal
distributions is $\sigma^2 = 64$ or $\sigma = 64.$ Common sense
says it should be the former so that $\sigma = 8.$ Even so, there
is poor separation between the two normal curves of test scores.
So we cannot expect both high sensitivity and high specificity.

Another difficulty is to determine what cutoff thyroid score separates people deemed to test 'positive' for thyroid disease
(higher value) from those testing 'negative'. We need to choose
this score in a way so that prevalence is 2%.
The easy path is to find the 98th percentile of $Norm(3, 8).$
In essence, this means to equate values of sensitivity and prevalence. In R, we compute the cutoff score to be 19.43.
 qnorm(.98,3,8)
 ## 19.42999

But this neglects that out population is a $mixture$ of people
with and without thyroid disease. We can simulate
test scores of this mixed population of people who have the disease
and those who don't. Then see what cutoff score is consistent
with 2% prevalence of the disease. From this point of view,
the cutoff value should be 19.83. (Throwing away the part of
the population that were simulated with nonsensical negative
scores, the cutoff is 21.28.)
  B = 10^6;  bin = rbinom(B, 1, .02)
  pop = bin*rnorm(B,11,8) + (1-bin)*rnorm(B,3,8)
  quantile(pop, .98) 
  ##      98% 
  ##      19.83799 

I  use the valus 19.8 to find sensitivity and specificity.
It is marked by the vertical line in the figure below.

Now, $SE = P(X > 19.8|\mu = 11) = .136.$
 SE = 1 - pnorm(19.8, 11, 8); SE
 ## 0.1356661

And $SP = P(0 < X < 19.8 | \mu = 3) = .628.$
 SP = diff(pnorm(c(0, 19.8), 3, 8)); SP
 ## 0.6283053

As noted above, both normal curves have substantial
probabilities below 0. Perhaps we should take the normal curves
as strictly positive and adjust the positive parts to enclose
unit probability for each curve. Then SE and (especially) SP become
larger:  $SE_{adj} = .148,\; SP_{adj} = .972.$
 SE.adj = SE/(1 - pnorm(0, 11, 8));  SE.adj
 ## 0.1481986
 SP.adj = SP/(1 - pnorm(0, 3, 8)); SP.adj
 ## 0.9723534

Finally, $PPV = P(\text{Has Disease}|\text{Positive Test}).$
According to your formula, this gives 0.0074 without adjustment
for negative areas, and 0.099 with adjustment. Both values are
quite low, but low PPV values are typical with low disease
prevalence. 
 PPV = .02*SE/(.02*SE + .98*(1-SP)); PPV
 ## 0.007393768
 PPV.adj = .02*SE.adj/(.02*SE.adj + .98*(1-SP.adj)); PPV.adj
 ## 0.09860948

Summary: As you can see, the unresolved questions how to
find a cutoff score and how to deal with the fact that
both distributions stray far into negative territory have
led to many possible paths, some taken here and some not.
I do not claim any of this is $the$ answer to your question.
When you figure out how to resolve the difficulties, I hope
this will help you pick a useful path to the answer you need.
