Question

CNNBC recently reported that the mean annual cost of auto insurance is 986 dollars. Assume the standard deviation is 216 dollars. You take a simple random sample of 98 auto insurance policies.

Find the probability that a single randomly selected value is less than 996 dollars.
P(X<996)=
Find the probability that a sample of size n = 98 is randomly selected with a mean less than 996 dollars.
P(M-996)=

Answer 1

To find the probability that a single randomly selected value is less than 996 dollars, we calculate the z-score and use the standard normal distribution. The probability is approximately 0.5181, or 51.81%. To find the probability that a sample of size n = 98 is randomly selected with a mean less than 996 dollars, we use the Central Limit Theorem. The probability is approximately 0.6736, or 67.36%.

Step-by-step explanation:

To find the probability that a single randomly selected value is less than $996, we can standardize the value and use the standard normal distribution. The formula for standardizing a value is: z = (x - mean) / standard deviation.

Using the given information:

• Mean (μ) = $986
• Standard deviation (σ) = $216
• Value (x) = $996

Substituting the values into the formula: z = (996 - 986) / 216 = 0.0463

Using a standard normal distribution table or a calculator, we can find the probability corresponding to a z-score of 0.0463. The probability is approximately 0.5181, or 51.81%.

Therefore, the probability that a single randomly selected value is less than $996 is 0.5181, or 51.81%.

To find the probability that a sample of size n = 98 is randomly selected with a mean less than $996, we can use the Central Limit Theorem. The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size increases.

The mean of the sample means (μM) is equal to the population mean (μ) which is $986, and the standard deviation of the sample means (σM) is obtained by dividing the population standard deviation (σ) by the square root of the sample size (√n).

Using the given information:

• Mean (μ) = $986
• Standard deviation (σ) = $216
• Sample size (n) = 98

Substituting the values into the formula: σM = σ / √n = $216 / √98 = $21.8404

Next, we can standardize the value using the formula z = (x - μM) / σM where x is the value of $996.

Substituting the values into the formula: z = (996 - 986) / 21.8404 = 0.4576

We can find the probability corresponding to a z-score of 0.4576 using a standard normal distribution table or a calculator. The probability is approximately 0.6736, or 67.36%.

Therefore, the probability that a sample of size n = 98 is randomly selected with a mean less than $996 is 0.6736, or 67.36%.