Answer to Question #160709 in Statistics and Probability for Soumyo

Question

Cannon shell impact position, as measured along the line of fire from the target point, can be described by a normal random variable X. It is found that 15.15% of shells fall 11.2 m or farther from the target in a direction toward the cannon, while 5.05% fall farther than 95.6 m beyond the target. what are the mean and standard deviation of distribution of X?

Answer 1

Let $X=$ the cannon shell impact position, as measured along the line of fire from the target point: $X\sim N(\mu, \sigma).$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0,1)$

$$P(X<-11.2)=P(Z<\dfrac{-11.2-\mu}{\sigma})=0.1515$$

$$\dfrac{-11.2-\mu}{\sigma}\approx-1.03002$$

$$P(X>95.6)=1-P(X\leq95.6)=$$

$$=1-P(Z\leq\dfrac{95.6-\mu}{\sigma})=0.0505$$

$$P(Z\leq\dfrac{95.6-\mu}{\sigma})=0.9495$$

$$\dfrac{95.6-\mu}{\sigma}\approx1.64002$$

$$95.6-1.64002\sigma=-11.2+1.03002\sigma$$

$$2.67004\sigma=106.8$$

$$\sigma=40$$

$$\mu=95.6-1.64002(40)=30$$

$\mu=30\ m,\sigma=40\ m$