# A distribution of values is normal with a mean of 203.8 and a standard deviation of 50.2.Find the probability that a randomly selected value is less than 113.4.P(X < 113.4) =?

A distribution of values is normal with a mean of 203.8 and a standard deviation of 50.2.
Find the probability that a randomly selected value is less than 113.4.
$P\left(X<113.4\right)=$

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Theodore Schwartz

Step 1
Solution:
Let X be the values.
From the given information, X follows normal distribution with mean $\mu =203.8$ and a standard deviation is $\sigma =50.2$.
Step 2
Then, the probability that a randomly selected value is less than 113.4 is
$P\left(X<113.4\right)=P\left(\frac{X-\mu }{\sigma }<\frac{113.4-203.8}{50.2}\right)$
$=P\left(Z<-1.801\right)$

Thus, the probability that a randomly selected value is less than 113.4 is 0.0359.

Jeffrey Jordon