Question

CNBC recently reported that the mean annual cost of auto insurance is $1027. Assume the distribution of auto insurance follows a normal distribution with a standard deviation of $228.

A) Find the probability that a single randomly selected value is less than $964. P(X < 964) = ?
B) Find the probability that a sample of size 67 is randomly selected with a mean less than $964. P( X < 964) = ?

Answer 1

To calculate the probability of a single insurance cost being less than $964, we convert the value to a z-score and refer to the standard normal distribution. For the probability of the sample mean being less than $964, we calculate the standard error, convert the sample mean to a z-score and consult the standard normal distribution.

Step-by-step explanation:

The question asked pertains to the probability that a single randomly selected auto insurance cost is less than $964 and the probability that the sample mean of 67 auto insurance costs is less than $964, given that the mean and standard deviation of the auto insurance costs are $1027 and $228 respectively. To solve this, we use the standard normal distribution for the first part and the sampling distribution of the sample mean for the second part.

Part A: Probability of a Single Value Being Less Than $964

First, we convert $964 to a z-score using the formula:

Z = (X - μ) / σ

Where X is the value of $964, μ is the mean of $1027, and σ is the standard deviation $228.

Next, we look up the z-score on the standard normal distribution table to find the probability P(X < 964).

Part B: Probability of a Sample Mean Being Less Than $964

For this part, we need to calculate the standard error of the mean which is σ/√(n), where n is the sample size, which in this case is 67. Then, we find the z-score for the sample mean the same way,

Z = (X-bar - μ) / (Standard Error)

We look up this z-score on the standard normal distribution table to find P( X < 964).