Answer to Question #171183 in Statistics and Probability for clyde

Question

 A normally distributed random variable X has a mean of 500 and a standard deviation of 40.

Answer 1

Q: P(X>530)

Mean($\mu$ )= 500

Standard deviation($\sigma$ )= 40

Normal distribution formula$\Rightarrow z=\dfrac{\bar x-\mu}{\sigma}$

$z=\dfrac{530-500}{40}= \dfrac{30}{40}=0.75$

Now, P(z>0.75)=0.5-0.2734=0.2266

Hence, P(X>530)=0.2266