The Z-score of a random variable X is defined as follows:

\(\displaystyle{Z}=\frac{{{X}-\mu}}{\sigma}\)

Here, \(\displaystyle\mu{\quad\text{and}\quad}\sigma\) 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,

\(\displaystyle{P}{\left({X}{<}{969}\right)}={P}{\left({\frac{{{\left({X}-\mu\right)}}}{{\sigma}}}{<}{\frac{{{969}-{954}}}{{{271}}}}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}{0.06}\right)}={0.5221}\)</span> (Using standard normal table: \(\displaystyle{P}{\left({Z}{<}{0.06}\right)}={0.5221}\)</span>)

Therefore, the probability that a single randomly selected value is less than 969 dollars is 0.5221.

\(\displaystyle{Z}=\frac{{{X}-\mu}}{\sigma}\)

Here, \(\displaystyle\mu{\quad\text{and}\quad}\sigma\) 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,

\(\displaystyle{P}{\left({X}{<}{969}\right)}={P}{\left({\frac{{{\left({X}-\mu\right)}}}{{\sigma}}}{<}{\frac{{{969}-{954}}}{{{271}}}}\right)}\)</span>

\(\displaystyle={P}{\left({Z}{<}{0.06}\right)}={0.5221}\)</span> (Using standard normal table: \(\displaystyle{P}{\left({Z}{<}{0.06}\right)}={0.5221}\)</span>)

Therefore, the probability that a single randomly selected value is less than 969 dollars is 0.5221.