The distribution of scores on an exam approximately follows a normal distribution N (14,4). What is the probability that a student who took that exam scored less than 12? What is the maximum score that a student must have followed on the exam to belong to the group of the 2.3% of the worst-scoring students?

Part 1) What is the probability that a student who took that exam scored less than 12?

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The notation N (14,4) means we have a mean of mu = 14 and standard deviation of sigma = 4 when talking about the normal distribution.

Let's convert the raw score x = 12 to its corresponding z score.

z = (x-mu)/sigma

z = (12-14)/4

z = -2/4

z = -0.5

The task of finding is identical to finding

Use a Z table and you should get roughly 0.3085

 0.3085

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Part 2) What is the maximum score that a student must have followed on the exam to belong to the group of the 2.3% of the worst-scoring students?

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We need to find a value of k such that

If you were to use a table, then you'd find that

Using a calculator, the value is k = -1.99539 approximately.

We'll plug this as the z value to solve for x.

z = (x-mu)/sigma

-1.99539 = (x-14)/4

-1.99539*4 = x-14

-7.98156 = x-14

x-14 = -7.98156

x = -7.98156+14

x = 6.01844

Assuming scores are given to the nearest whole number, then we'd round that x value to x = 6. This tells us that

If a student scores a 6, then they rank at the very top of the "worst 2.3%" group.

r(0,90)(x,y) . t(-5,0)

last option

step-by-step explanation:

∆def to ∆d"e"f"

- rotate 90 counterclockwise degrees

- then translated 5 to the left (x --> x - 5)

answer is the last option