Let X denote the number of scores in a test. If X is normally distributed with mean 100 and standard deviation 15, then the probability that X does not exceed 130 is,
(Given \( P(0\le Z\le 2)=0.4772 \), where \( Z \) be the standard normal variate)
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 0.9772
Let Z be the standard normal variate for X:
\( \displaystyle Z=\frac{X-\mu}{\sigma}=\frac{130-100}{15}=2 \)
“Does not exceed 130” means \( X\le 130 \), i.e. \( Z\le 2 \). The standard normal curve is symmetric about 0, so \( P(Z\le 0)=0.5 \):
\( \displaystyle P(Z\le 2)=P(Z\le 0)+P(0\le Z\le 2) \)
\( \displaystyle =0.5+0.4772=0.9772 \)
Why the other options are incorrect
Option (1) 0.4772 — Incorrect. 0.4772 is only \( P(0\le Z\le 2) \); the half below the mean, 0.5, must be added.
Option (2) 0.5 — Incorrect. 0.5 is \( P(Z\le 0) \), the probability of not exceeding the mean 100.
Option (4) 0.0228 — Incorrect. 0.0228 is \( P(Z\gt 2)=0.5-0.4772 \), the probability that X exceeds 130.