An automobile in surer has found that repair claims have a mean of $1520 and a stardard deviation of $770. Suppose that the next 100 claims can be ragarded as a random sample from the long-run claims process
1. What is the mean and standrd deviation of the average x(bar) of the next 100 claims?
2. What is the probability that the average x (bar) of the 100 claims in less than $1500?
(Please explain the steps taken so I can understand)
To find the mean (μ) and standard deviation (σ) of the average (x-bar) of the next 100 claims, you can use the following formulas:
1. Mean (μ) of the average (x-bar):
The mean of the average is equal to the mean of the original data. Therefore, the mean (μ) of the average (x-bar) of the 100 claims is also $1520.
2. Standard deviation (σ) of the average (x-bar):
The standard deviation of the average is equal to the standard deviation of the original data divided by the square root of the sample size (n). Therefore, the standard deviation (σ) of the average (x-bar) of the 100 claims is calculated as follows:
σ = σ_original / √n
Given that the standard deviation (σ_original) of the original data is $770 and the sample size (n) is 100, we can calculate the standard deviation (σ) of the average as:
σ = $770 / √100
σ = $770 / 10
σ = $77
So, the mean (μ) of the average (x-bar) is $1520 and the standard deviation (σ) of the average (x-bar) is $77.
To find the probability that the average (x-bar) of the 100 claims is less than $1500, we need to calculate the z-score and then find the corresponding probability using the standard normal distribution.
The z-score is calculated using the formula:
z = (x - μ) / σ
where x is the value we are interested in (in this case, $1500), μ is the mean ($1520), and σ is the standard deviation ($77).
So,
z = ($1500 - $1520) / $77
z = -20 / $77
z ≈ -0.26
Now, we can use a standard normal distribution table or a calculator to find the probability associated with the z-score of -0.26. The probability of the average (x-bar) of the 100 claims being less than $1500 is the area under the curve to the left of the z-score, which can be found from the standard normal distribution table or a calculator.
In this case, from the standard normal distribution table, we can find that the probability (P) associated with a z-score of -0.26 is approximately 0.397.
Therefore, the probability that the average (x-bar) of the 100 claims is less than $1500 is 0.397, or approximately 39.7%.