 # CNNBC recently reported that the mean annual cost of auto insurance is 954 dollars. Assume the standard deviation is 271 dollars. Tabansi 2021-03-02 Answered

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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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.

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