Name: Danielle

Who is asking: Student
Level: Secondary

Question: A teacher gave a test on which the students' marks were normally distributed, but the results were pathetic. The mean was 52% and the standart deviation was 12%. The teacher decided that the top 10% of the students should get A's, the next 20% should get B's, the next 40% should get C's, the next 20% should get D's, and the bottom 10% should get F's. To the nearest percent, what are the cutoff marks that will result in an A, B, C, D, and F?

Could you please explain how to do this question for me, and what formula's you used. I am a little confused!

Thank you!

Hi Danielle,

As with many problems that involve the normal distribution, the first task is to transform to the standard normal distribution. The transformation from a normal distribution (X) with mean mu and standatd distribution sigma, to the standard normal distribution (Z) is

Z = (X - mu)/sigma Your situation has mu = 52% and sigma = 12% so the transformation is
Z = (X - 52)/12

The teacher wants 10% of the students to receive A's. From a table for the standard normal distribution, 10% of the area lies above 1.282. Thus the grade that separates the A's from the B's is the number X that satisfies

1.282 = (X - 52)/12 That is X = 67.4

The next 20% of the students are to receive B's. From the normal table 30% of the area lies above 0.524. (This is the division between B's and C's so 30% of the students have grades that lie above this number.) Thus

0.524 =  (X - 52)/12 and thus X = 58.3

You should now be able to find the remaining division points.

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