Question

A test is given and the average (μX) score out of 100 was only a 53.1, with a SD (σX) of 8.9. Assuming the grades followed a normal distribution, use the Z table or Excel and formulas to find the dividing line (test scores) between the A's, B's, C's, D's, and E's. Starting from the top, the teacher will give the highest 15% A's, the next 10% B's, the next 30% C's, the next 25% D's, and the bottom 20% will receive E's.

Answer 1

Given = 53.1, = 8.9 .

To find the probability, we need to find the z scores.

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Those who get A's = Top 15%. Therefore there are 85% below them.

P(X < x) = 0.85

The z score at 0.85 is = 1.0364

Therefore 1.0364 = (X - 53.1) / 8.9

Solving for X, we get, X = (1.0364 * 8.9) + 53.1

X = 62.32

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Those who get B's = Next 10%. To get a B, you have to be in the top (15 +10) = 25%

Therefore there are 75% below them.

P(X < x) = 0.75

The z score at 0.75 is = 0.6745

Therefore 0.6745 = (X - 53.1) / 8.9

Solving for X, we get, X = (0.6745 * 8.9) + 53.1

X = 59.1

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Those who get C's = Next 30%. To get a C, you have to be in the top (15 + 10 + 30) = 55%

Therefore there are 45% below them.

P(X < x) = 0.45

The z score at 0.45 is = -0.1257

Therefore -0.1257 = (X - 53.1) / 8.9

Solving for X, we get, X = (-0.1257 * 8.9) + 53.1

X = 51.98

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Those who get D's = Next 25%. To get a C, you have to be in the top (15 + 10 + 30 + 25) = 80%

Therefore there are 20% below them.

P(X < x) = 0.20

The z score at 0.2 is = -0.8416

Therefore -0.8416 = (X - 53.1) / 8.9

Solving for X, we get, X = (-0.8416 * 8.9) + 53.1

X = 45.61

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Therefore

The score to get an A = 62.32 and Above

The score to get a B = 59.1 and above but less than 62.32

The score to get a C = 51.98 and above but less than 59.1

The score to get a D = 45.61 and above but less than 51.98

The score to get an E is any score less than 45.61 i.e = 45.6