Let X be a normal random variable with mean 12 and variance 4. Find the value of c such that P{X>c}=.10.

Let X be a normal random variable with mean 12 and variance 4. Find the value of c such that P{X>c}=.10.
You can still ask an expert for help

Want to know more about Probability?

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Aamina Herring
Given: X is normal random variable with mean $\mu =12$ variance ${\sigma }^{2}=4$
To find: Value of c such that P[X>c]=0.10
Solution: We have
P[X>c]=0.1
$⇒P\left[X\le c\right]=1-0.1=0.9$
$⇒P\left[\frac{X-\mu }{\sigma }\le \frac{c-12}{2}\right]=0.9$
$⇒P\left[Z\le \frac{c-12}{2}\right]=0.9$
$⇒\mathrm{\Phi }\left(\frac{c-12}{2}\right)=0.9$
$⇒\frac{c-12}{2}=1.28$
$⇒c-12=2.56$
$⇒c=14.56$