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QUESTION 23 Suppose a certain college has its own entrance exam, and scores on this exam follow a normal distribution with me
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Answer #1

Let X denotes the scores of a certain college.

X ~ N(\small \mu=150, 0 =20)

Then,

- N(0,1) o

Let Z X- T

Therefore, P(Z <) = (2) and (-2) =1-02)

Bob's score was 130.

We have to find P(X>130),

PX > 130) = 1- P(X < 130

                       X- = 1-PC VI 130 – 150 20 o

                       =1-PZ-1

                       =1 - Φ(-1) =1-1+ Φ(1)

                                                = Φ(1) = 0.8413-        (Value from z table)

The percentage of the students who had scores that fell above Bob's scores is 84% (Option 3).

z table :

Z 0.0 0.1 02 03 0.4 05 0.6 0.7 0.8 0.9 1.0 1.1 12 13 14 15 1.6 17 18 1.9 2.0 21 22 23 2.4 25 2.6 2.7 2.8 2.9 3.0 3.1 3.2 3.3

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