Question

Problem 9: Given X is a random variable with normal distribution, N (?, ?2). Given ?-6 and ?2-0.25, determine b, c, and d such that: I. P (Xs b) - 0.05 Il. P (X2 c) 0.40

i need this step by step please

Answer 1

Given : $X\to N(\mu=6,\sigma^2=0.25 )$

Therefore , $\sigma=\sqrt{0.25}=0.5$

I )

$P(X\leq b)=0.05$

$P(X\leq -b)=0.05$

$P(\frac{X-\mu}{\sigma}\leq -\frac{b-6}{0.5})=0.05$

$P(Z\leq -\frac{b-6}{0.5})=0.05$

From statistical table ,

$\therefore -\frac{b-6}{0.5}=1.64$

$\therefore -b=1.64*0.5-6=-5.18$

$\therefore b=5.18$

II )

$P(X\geq c)=0.40$

$P(\frac{X-\mu}{\sigma}\geq \frac{c-6}{0.5})=0.40$

$P(Z\geq \frac{c-6}{0.5})=0.40$

From statistical table ,

$\therefore \frac{c-6}{0.5}=0.25$

$\therefore c=0.25*0.5+6=6.1250$

III )

$P(|X-6|\leq d)=0.95$

$P(-d<X-6<d)=0.95$

$P(-\frac{d}{0.5}<\frac{X-6}{0.5}<\frac{d}{0.5})=0.95$

$P(-\frac{d}{0.5}<Z<\frac{d}{0.5})=0.95$

$P(Z>-\frac{d}{0.5})-P(Z>\frac{d}{0.5})=0.95$

$1-P(Z<-\frac{d}{0.5})-P(Z>\frac{d}{0.5})=0.95$

$1-P(Z>\frac{d}{0.5})-P(Z>\frac{d}{0.5})=0.95$

$1-2*P(Z>\frac{d}{0.5})=0.95$

$2*P(Z>\frac{d}{0.5})=1-0.95=0.05$

$P(Z>\frac{d}{0.5})=0.05/2=0.025$

From statistical table ,

$\therefore \frac{d}{0.5}=1.96$

$\therefore d=1.96*0.5=0.98$