Question

Suppose x is a normally distributed random variable with μ = 50 and Ϭ = 3. Find a value of the random variable, call it x0, such that a. P(x ≤ x0) = 0.8413 b. P(x > x0) = 0.25 c. P(x > x0) = 0.95 d. P(41 ≤ x < x0) = 0.8630 e. 10% of the values of x are less than x0. f. 1% of the values of x are greater than x0.

Answer 1

$$x\sim N(\mu, \sigma^2)$$

Given $\mu=50, \sigma=3$

a.

$$P(x\leq x_0)=0.8413$$

$$P(x\leq x_0)=P(z\leq z_0 )=0.8413$$

$$z_0=0.999815={x_0-\mu \over \sigma}$$

$$x_0=3(0.999815)+50\approx53$$

b.

$$P(x>x_0)=0.25$$

$$P(x\leq x_0)=1-P(x>x_0)=1-0.25=0.75$$

$$P(x\leq x_0)=P(z\leq z_0 )=0.75$$

$$z_0=0.674490={x_0-\mu \over \sigma}$$

$$x_0=3(0.674490)+50\approx52$$

c.

$$P(x>x_0)=0.95$$

$$P(x\leq x_0)=1-P(x>x_0)=1-0.95=0.05$$

$$P(x\leq x_0)=P(z\leq z_0 )=0.05$$

$$z_0=-1.645={x_0-\mu \over \sigma}$$

$$x_0=3(-1.645)+50\approx45$$

d.

$$P(41\leq x< x_0)=P(x<x_0)-P(x\leq 41)=$$

$$=P(z<z_0)-P(z\leq{41-50 \over 3})=$$

$$=P(z<z_0)-P(z\leq-3)=$$

$$=P(z<z_0)-0.00135=0.8630$$

$$P(z<z_0)=0.86435$$

$$z_0=1.1={x_0-\mu \over \sigma}$$

$$x_0=3(1.1)+50\approx53.3$$

e.

$$P(x< x_0)=0.1$$

$$P(x< x_0)=P(z<z_0)=0.1$$

$$z_0=-1.28155={x_0-\mu \over \sigma}$$

$$x_0=3(-1.28155)+50\approx46.155$$

f.

$$P(x>x_0)=0.01$$

$$P(x\leq x_0)=1-P(x>x_0)=1-0.01=0.99$$

$$P(x\leq x_0)=P(z\leq z_0 )=0.99$$

$$z_0=2.326={x_0-\mu \over \sigma}$$

$$x_0=3(2.326)+50\approx56.978$$