(a) To find the probability that a single claim, chosen at random, will be less than $680, we need to calculate the z-score and use the standard normal distribution table.
First, we calculate the z-score using the formula:
z = (x - μ) / σ
where:
x = $680 (the value we want to find the probability for)
μ = $710 (mean of the repair claims)
σ = $660 (standard deviation of the repair claims)
Plugging in the values, we get:
z = (680 - 710) / 660
z ≈ -0.045
Now, we can use the standard normal distribution table (or a calculator) to find the probability corresponding to this z-score. The table gives us the probability of getting a value less than the z-score, so we need to look up the z-score -0.045.
The probability is approximately 0.4832.
Therefore, the probability that a single claim, chosen at random, will be less than $680 is approximately 0.4832.
(b) Since the next 50 claims can be regarded as a random sample, we can use the Central Limit Theorem to approximate the distribution of the sample mean.
According to the Central Limit Theorem, the distribution of the sample mean is approximately Normal with a mean equal to the population mean (μ = $710) and a standard deviation equal to the population standard deviation divided by the square root of the sample size (σ/√n = $660/√50).
Now, we need to find the probability that the average x¯ of the 50 claims is smaller than $680. We can use the same formula as in part (a) to calculate the z-score:
z = (x - μ) / (σ/√n)
Plugging in the values, we get:
z = (680 - 710) / (660/√50)
z ≈ -1.646
Using the standard normal distribution table (or a calculator), we can find the probability corresponding to this z-score. The probability is approximately 0.0505.
Therefore, the probability that the average x¯ of the 50 claims is smaller than $680 is approximately 0.0505.
(c) If a sample larger than 50 claims is considered, there would be less chance of getting a sample with an average smaller than $680.