Hi everyone
The personnel department of a large corporation gives two aptitude
tests to job applicants. One measures verbal ability; the other,
quantitative ability. From many years? experience, the company has
found that the verbal scores tend to be normally distributed with a
mean of 50 and a standard deviation of 10. The quantitative scores are
normally distributed with a mean of 100 and a standard deviation of
20, and they appear to be independent of the verbal scores. A
composite score, C, is assigned to each applicant, where
C = 3(verbal score) + 2(quantitative score).
If company policy prohibits hiring anyone whose C score is below 375,
what percentage of applicants will be summarily rejected?

Hi, chanchai-ga:

Let V = (verbal score) and Q = (quantitative score) be normally
distributed random variables, as you have described.

Their linear combination:

C = 3V + 2Q

is (perhaps surprisingly) again normally distributed.  The mean of C
is easy to work out, mainly because the "expectation" is a linear
functional:

E(C) = 3E(V) + 2E(Q)

where E( ) denotes the expected value or mean of the random variable.  

Since the mean verbal score is assumed to be 50 and the mean
quantitative score 100, the mean composite score works out to 3*50 +
2*100 = 350.

Once we determine the standard deviation of the composite score, we
will know everything we need to about C.  It'll be like using the
Pythagorean theorem; see the formulas here:

[Normal Sum Distribution -- from MathWorld]
http://mathworld.wolfram.com/NormalSumDistribution.html

The variance is the square of the standard deviation, and as you'll
note from the formula given at the Web page linked above, the variance
of C is then the sum of the variance of 3V and the variance of 2Q.  So
the standard deviation of C, sigma_C, satisfies:

sigma_C = SQUARE_ROOT( (sigma_{3V})^2 + (sigma_{2Q})^2 )

Now the standard deviation of 3V is three times the standard deviation
of V, and likewise the standard deviation of 2Q is twice that of Q. 
So:

sigma_C = SQUARE_ROOT( (30)^2 + (40)^2 )

        = 50

Your Question asks what percentage of applicants will be rejected,
based as it seems on a normal distribution for those falling below 375
in their composite scores.  Since the mean of C is 350, we can see
already that more than half of those who are tested will "fail".  The
passing mark set at 375 is one-half standard deviation above the mean,
i.e. 375 - 350 = 25 is half of the standard deviation 50 found above.

From a "left tail" (area of the standard normal distribution below a
multiple of the standard deviation) normal distribution table, such as
given here:

[LEFT TAIL AREAS OF THE STANDARD NORMAL DISTRIBUTION]
http://www.wlu.ca/~wwwmath/courses/cou100/ma141/z_table.pdf

we learn that the fraction of those whose scores are less than:

375 = mean + 0.50 * sigma

will be A(0.50) = 0.691.  That is, 69.1% of all applicants will be
rejected on the basis of their composite scores.  (Tough policy!)

[Note that by modelling the distribution of scores with a continuous
distribution (in this the normal distribution), we treat the chance of
scoring _exactly_ 375 as if it were zero.  So we don't have to quibble
about whether someone falling on the boundary exactly would pass or
fail; the fraction of such cases is considered negligible.  Modelling
with a discrete distribution would force us to be a bit more careful
to distinguish scores "below 375" from ones "less than or equal to
375".]

Please let me know by requesting Clarification if some additional
explanation would be helpful in understanding how this Answer was
derived.

regards, mathtalk-ga