Suppose that X is a normal random variable with mean 5. If P{X&gt;9}=.2, approximately what is Var(X)?

X is an averaged, normal random variable.&nbsp;μ=5&nbsp;and&nbsp;P[X&gt;9]=0.2Let X's variance be&nbsp;σ2&nbsp;then.,P[X&gt;9]=0.2⇒P[X≤9]=1−0.2=0.8⇒P[x−μσ≤9−4σ]=0.8⇒P[Z≤4σ]=0.8⇒ϕ(4σ)=0.8⇒4σ=0.845⇒σ=40.845=4.73Therefore, X's variance isvar(X)=σ2=(4.73)2=22.37=22

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Answered question

2021-11-10

Suppose that X is a normal random variable with mean 5. If P{X>9}=.2, approximately what is Var(X)?

Answer & Explanation

Ched1950

Beginner2021-11-11Added 21 answers

X is an averaged, normal random variable.

$μ = 5$ and $P [X > 9] = 0.2$
Let X's variance be $σ^{2}$ then.,
$P [X > 9] = 0.2$
$\Rightarrow P [X \leq 9] = 1 - 0.2 = 0.8$
$\Rightarrow P [\frac{x - μ}{σ} \leq \frac{9 - 4}{σ}] = 0.8$
$\Rightarrow P [Z \leq \frac{4}{σ}] = 0.8$
$\Rightarrow ϕ (\frac{4}{σ}) = 0.8$
$\Rightarrow \frac{4}{σ} = 0.845$
$\Rightarrow σ = \frac{4}{0.845} = 4.73$
Therefore, X's variance is

$v a r (X) = σ^{2} = {(4.73)}^{2} = 22.37 = 22$

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