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(x<969)=P(x<969)=? Write your answers as numbers accurate to 4 decimal places.

Question
Random variables
asked 2021-03-02
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.
\(\displaystyle{P}{\left({x}{<}{969}\right)}={P}{\left({x}{<}{969}\right)}=\)</span>?
Write your answers as numbers accurate to 4 decimal places.

Answers (1)

2021-03-03
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.
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