An automobile insurer has found that repair claims have a mean of $920 and a standard deviation of $870. Suppose that the next 100 claims can be regarded as a random sample from the long-run claims process. The probability that the average, x bar, of these 100 claims is larger than $1,000 is approximately
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http://davidmlane.com/hyperstat/z_table.html
To calculate the probability that the average, x bar, of the 100 claims is larger than $1,000, we can use the sample mean and the Central Limit Theorem.
The Central Limit Theorem states that the sampling distribution of the sample mean approaches a normal distribution as the sample size gets larger, regardless of the shape of the population distribution.
In this case, we know that the population mean, µ, is $920 and the population standard deviation, σ, is $870. Since we have a large sample size of 100 claims, we can assume that the sample means will follow a normal distribution.
To calculate the probability, we need to convert the sample mean to a standard score (z-score) using the formula:
z = (x - µ) / (σ / √n)
where x is the sample mean, µ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x = $1,000, µ = $920, σ = $870, and n = 100.
Plugging in the values, we get:
z = (1000 - 920) / (870 / √100)
= 80 / (870 / 10)
= 80 / 87
Now, we can find the probability by looking up the z-value in the standard normal distribution table. The z-value of 80 / 87 is approximately 0.9195.
So, the probability that the average, x bar, of these 100 claims is larger than $1,000 is approximately 1 - 0.9195 = 0.0805, or 8.05%.