Question

Given that a random variable X is normally distributed with a mean of 2 and a variance of 4, find the value of x such that P(X < x)

Answer 1

Given that a random variable X is normally distributed with a mean of 2 and a variance of 4, find the value of x such that P(X < x)=0.99   using the cumulative standard normal distribution table

Answer:

6.642

Step-by-step explanation:

Given that mean = 2

standard deviation = 2

Let X be the random Variable

Then X $\sim$ N(n,$\sigma$)

X $\sim$ N(2,2)

By Central limit theorem;

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

$z = \dfrac{X - 2}{2} \sim N(0,1)$

P(X<x) = 0.09

$P(Z < \dfrac{X-\mu}{\sigma })= 0.99$

$P(Z < \dfrac{X-2}{2})= 0.99$

P(X < x) = 0.99

$P(\dfrac{X-2}{2}< \dfrac{X-2}{2})=0.99$

$P(Z< \dfrac{X-2}{2})=0.99$

$\phi ( \dfrac{X-2}{2})=0.99$

$( \dfrac{X-2}{2})= \phi^{-1} (0.99)$

$( \dfrac{X-2}{2})= 2.321$

X -2 = 2.321 × 2

X -2 = 4.642

X = 4.642 +2

X = 6.642