 Tabansi

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.
$P\left(x<969\right)=P\left(x<969\right)=$? Corben Pittman

The Z-score of a random variable X is defined as follows:
$Z=\frac{X-\mu }{\sigma }$
Here,  are the mean and standard deviation of X, respectively.
Take into account the X random variable, which represents the annual cost of car insurance.
The information provided indicates that X has a normal distribution, with a mean of 954 and a standard deviation of 271. The sample size, n is 96.
The likelihood that any given value chosen at random 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: $P\left(Z<0.06\right)=0.5221$)
Thus, there is a 0.5221 percent chance that a single randomly chosen value will be less than 969 dollars. Jeffrey Jordon