# CNNBC recently reported that the mean annual cost of auto insurance is 954 dollars. Assume the standard deviation is 271 dollars.

CNNBC recently reported that the mean annual cost of auto insurance is 954 dollars. Assume the standard deviation is 271 dollars. You take a simple random sample of 96 auto insurance policies.
Find the probability that a single randomly selected value is less than 969 dollars.
$P\left(x<969\right)=P\left(x<969\right)=$?

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Corben Pittman

The Z-score of a random variable X is defined as follows:
$Z=\frac{X-\mu }{\sigma }$
Here, $\mu \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\sigma$ are the mean and standard deviation of X, respectively.
Consider a random variable X that defines the annual cost of auto insurance.
According to the given information X follows normal distribution with mean 954 and standard deviation 271. The sample size, n is 96.
The probability that a single randomly selected value is less than 969 dollars is

$P\left(X<969\right)=P\left(\frac{\left(X-\mu \right)}{\sigma }<\frac{969-954}{271}\right)$

$=P\left(Z<0.06\right)=0.5221$

(Using standard normal table: $P\left(Z<0.06\right)=0.5221$)
Therefore, the probability that a single randomly selected value is less than 969 dollars is 0.5221.

Jeffrey Jordon