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 [tex]\sim[/tex] N(n,[tex]\sigma[/tex])
X [tex]\sim[/tex] N(2,2)
By Central limit theorem;
[tex]z = \dfrac{X - \mu}{\sigma} \sim N(0,1)[/tex]
[tex]z = \dfrac{X - 2}{2} \sim N(0,1)[/tex]
P(X<x) = 0.09
[tex]P(Z < \dfrac{X-\mu}{\sigma })= 0.99[/tex]
[tex]P(Z < \dfrac{X-2}{2})= 0.99[/tex]
P(X < x) = 0.99
[tex]P(\dfrac{X-2}{2}< \dfrac{X-2}{2})=0.99[/tex]
[tex]P(Z< \dfrac{X-2}{2})=0.99[/tex]
[tex]\phi ( \dfrac{X-2}{2})=0.99[/tex]
[tex]( \dfrac{X-2}{2})= \phi^{-1} (0.99)[/tex]
[tex]( \dfrac{X-2}{2})= 2.321[/tex]
X -2 = 2.321 × 2
X -2 = 4.642
X = 4.642 +2
X = 6.642
