# In a large section of a statistics​ class, the points for the final exam are normally​ distributed, with a mean of 72 and a standard deviation of 7. Grades are assigned such that the top​ 10% receive​ A's, the next​ 20% received​ B's, the middle​ 40% receive​ C's, the next​ 20% receive​ D's, and the bottom​ 10% receive​ F's. Find the lowest score on the final exam that would qualify a student for an​ A, a​ B, a​ C, and a D.The lowest score that would qualify a student for an A is________  (Round up to the nearest integer as​ needed.)The lowest score that would qualify a student for a B is_______  (Round up to the nearest integer as needed.)The lowest score that would qualify a student for a C is ______   (Round up to the nearest integer as needed.)The lowest score that would qualifty a student for a D is______  (  Round up to the nearesr integer as needed.)

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In a large section of a statistics class, the points for the final exam are normally distributed, with a mean of 72 and a standard deviation of 7. Grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40% receive C's, the next 20% receive D's, and the bottom 10% receive F's. Find the lowest score on the final exam that would qualify a student for an A, a B, a C, and a D.

The lowest score that would qualify a student for an A is________  (Round up to the nearest integer as needed.)

The lowest score that would qualify a student for a B is_______  (Round up to the nearest integer as needed.)

The lowest score that would qualify a student for a C is ______   (Round up to the nearest integer as needed.)

The lowest score that would qualifty a student for a D is______  (  Round up to the nearesr integer as needed.)

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Step 1

Finding the lowest score that would qualify a student for an A grade:

The points for the final exam are normally distributed with mean 72 and a standard deviation of 7. The top 10% (=0.10) of the students received an A grade. Thus, the lowest score that would qualify a student for an A grade is the upper limit of the area under the normal curve to the left of 0.90 (=1–0.10).

That is, to find the lowest score that would qualify a student for an A grade is the x-value corresponding to the probability 0.90 is obtained.

Software Procedure:

Step by step procedure to find the x-value by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter Mean as 72 and Standard deviation as 7.
• Choose probability and Left Tail for the region of the curve to shade.
• Enter the probability as 90.
• Click OK.

Output using MINITAB software is as follows:

Step 2

Thus, the lowest score that would qualify a student for an A grade is 81.

Step 3

Finding the lowest score that would qualify a student for a B grade:

The top 20% (=0.20) of the students received B grade and 10% (=0.10) received A grade. Thus, the lowest score that would qualify a student for a B grade is the upper limit of the area under the normal curve to the left of 0.70 (=1–0.30).

That is, to find the lowest score that would qualify a student for a B grade is the x-value corresponding to the probability 0.70 is obtained.

Software Procedure:

Step by step procedure to find the x-value by using MINITAB software is as follows:

• Choose Graph > Probability Distribution Plot > View Probability > OK.
• From ...

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