Answer to Question #160709 in Statistics and Probability for Soumyo

Question #160709

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?


1
Expert's answer
2021-02-04T05:33:07-0500

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

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


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

11.2μσ1.03002\dfrac{-11.2-\mu}{\sigma}\approx-1.03002



P(X>95.6)=1P(X95.6)=P(X>95.6)=1-P(X\leq95.6)=

=1P(Z95.6μσ)=0.0505=1-P(Z\leq\dfrac{95.6-\mu}{\sigma})=0.0505

P(Z95.6μσ)=0.9495P(Z\leq\dfrac{95.6-\mu}{\sigma})=0.9495

95.6μσ1.64002\dfrac{95.6-\mu}{\sigma}\approx1.64002

95.61.64002σ=11.2+1.03002σ95.6-1.64002\sigma=-11.2+1.03002\sigma

2.67004σ=106.82.67004\sigma=106.8

σ=40\sigma=40

μ=95.61.64002(40)=30\mu=95.6-1.64002(40)=30

μ=30 m,σ=40 m\mu=30\ m,\sigma=40\ m




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