Question

Arrange the probabilities for the following normal distribution problem from lowest to highest:
A. \(\mu=604,\ \sigma=50,\ P(X<635)\);
B. \(\mu=48,\ \sigma=10,\ P(X<20)\);
C. \(\mu=37,\ \sigma=5,\ P(X>35)\);
D. \(\mu=156,\ \sigma=10,\ P(X>170)\).

• A. \(A,D,B,C\)
• B. \(A,C,D,B\)
• C. \(B,D,C,A\)
• D. \(B,D,A,C\)

Hint:

For normal distribution, first convert to \(Z=\frac{X-\mu}{\sigma}\), then compare probabilities using the standard normal curve.

Answer 1

The Correct Option is C

Concept:
Convert each normal variable into standard normal form: \[ Z=\frac{X-\mu}{\sigma} \]

Step 1: For A. \[ Z=\frac{635-604}{50}=\frac{31}{50}=0.62 \] \[ P(X<635)=P(Z<0.62) \] This is a relatively large probability.

Step 2: For B. \[ Z=\frac{20-48}{10}=-2.8 \] \[ P(X<20)=P(Z<-2.8) \] This is very small.

Step 3: For C. \[ Z=\frac{35-37}{5}=-0.4 \] \[ P(X>35)=P(Z>-0.4) \] This is greater than \(0.5\).

Step 4: For D. \[ Z=\frac{170-156}{10}=1.4 \] \[ P(X>170)=P(Z>1.4) \] This is small, but greater than B.

Step 5: Arrange lowest to highest. \[ B<D<C<A \] Therefore: \[ B,D,C,A \] \[ \therefore \text{Correct Answer is (C)} \]