If \( X \) is normally distributed with mean 30 and standard deviation 5, the value of \( P(26\le X\le 34) \) is :(Where \( P(0\le Z\le 0.8)=0.2881 \))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.5762
Let \( Z=\frac{X-\mu}{\sigma} \), the standard normal variable, with \( \mu=30 \) and \( \sigma=5 \):
\( \displaystyle X=26:\ Z=\frac{26-30}{5}=-0.8 \)
\( \displaystyle X=34:\ Z=\frac{34-30}{5}=0.8 \)
\( \displaystyle P(26\le X\le 34)=P(-0.8\le Z\le 0.8) \)
The standard normal curve is symmetric about 0, so the area from −0.8 to 0 equals the area from 0 to 0.8:
\( \displaystyle P(-0.8\le Z\le 0.8)=2\,P(0\le Z\le 0.8) \)
\( \displaystyle =2\times 0.2881 \)
\( \displaystyle =0.5762 \)