SOLUTION: I really need help on this. Thanks so much. I would really appreciate it. Final averages are typically approximately normally distributed with a mean of 78 and a standard devia

Question 756218: I really need help on this. Thanks so much. I would really appreciate it.
Final averages are typically approximately normally distributed with a mean of 78 and a standard deviation of 11. Your professor says that the top 6% of the class will receive an A; the next 24%, a B; the next 41%, a C; the next 18%, a D; and the bottom 11%, an F. (Give your answers correct to one decimal place.)
(a) What average must you exceed to obtain an A?
(b) What average must you exceed to receive a grade of C or better?
(c) What average must you obtain to pass the course? (You'll need a D or better.)
You may need to use the appropriate table in Appendix B to answer this question.

Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Final averages are typically approximately normally distributed with a mean of 78 and a standard deviation of 11. Your professor says that the top 6% of the class will receive an A; the next 24%, a B; the next 41%, a C; the next 18%, a D; and the bottom 11%, an F. (Give your answers correct to one decimal place.)
----
Using a TI-84 to get these answers.
-----------------
(a) What average must you exceed to obtain an A?
Find the z-value with a left-tail of 94%
invNorm(0.94) = 1.5548
Using the formula z = (x-u)/s to solve for "x"
you get x = z*s+u
x is the score; z, s, and u you now know.
So solve for "x":
--------------------
score = 1.5548*11+78 = 95.1
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(b) What average must you exceed to receive a grade of C or better?
invNorm(0.29) = -0.5534
score = -9.5534*11+78 = 71.91
----------------------------------------

(c) What average must you obtain to pass the course? (You'll need a D or better.)
invNorm(-.11) = -1.2265
score = -1.2265*11+78 = 64.51
---------------
Cheers,
Stan H.
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