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It is given that $X \sim \mathcal{N}(1.5,(3.2)^2)$. Find the probability that the random chosen value of $X$ is less than $-2.4$.

Solution: $Pr(X<-2.4)=1-Pr(X<2.4)=1-Pr\left[Z < \cfrac{2.4-1.5}{3.2}\right]=1-Pr[Z < 0.28]=1-0.61026=0.389$

where $Z \sim \mathcal N(0,1)$

($0.61026$ value is obtained from the z table).

Is my answer right? It is multiple choice question and my answer doesn't match the options. Please correct me if wrong.

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2 Answers 2

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$P(X<-2.4)=P(\frac{X-1.5}{3.2} <\frac{-2.4-1.5}{3.2})=P(Z<-1.21875)\cong 1-0.88877 =0.11123$

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The problem with your solution is the first equality. It is not true that if $X \sim N(\mu, \sigma^2)$ , $P(X< -a) = 1-P(X<a)$ unless $X = Z\sim N(0,1),$ the "standard normal". As in Mike's answer, you must first convert your probability statement to a probability statement about $Z$ before flipping the sign..

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