Answer to Question #167214 in Statistics and Probability for Sab

Question

The random variable X has a normal distribution with mean 6.5 and variance 9. Find the value of x, call it

xo, such that P(xo ≤ x ≤ 7) = 0.6853

Answer 1

We have that:

$\mu = 6.5$

$var=\sigma^2=9\implies \sigma=\sqrt 9=3$

$P(x_o\le X\le7)=P(X\le7)-P(X\le x_o)$

$P(x_o\le X\le7)=P(X \leq 7)-P(X\leq x_0)=0.6853 \implies P(X\le x_o)=P(X\le7)-0.6853$

$P(X\le x)=P(Z\le\frac{x-\mu}{\sigma})$

$P(X\le7)=P(Z\le\frac{7-6.5}{3})=P(Z\le 1.67)=0.9525$

$P(X\le x_o)=P(X\le7)-0.6853=0.9525-0.6853=0.2672$

Z-value –0.62 of corresponds to 26.72% area under the curve. Then

$\frac{x_o-\mu}{\sigma}=\frac{x_o-6.5}{3}=-0.62\implies x_o=-0.62\cdot3+6.5=4.64$

Answer:4.64