The repair cost for a certain model of laptop computer has a normal distribution with a mean of $100, and with a standard deviation of $25. If 16 ovens are sent for repairs, find the probability that the average cost of repairs will be within $10 of the mean.
It would help if you proofread your questions before you posted them.
Have no data on ovens.
Z = (score-mean)/SEm
SEm = SD/√n
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to Z.
To find the probability that the average cost of repairs will be within $10 of the mean, we first need to calculate the standard error of the mean. The standard error of the mean (SE) is calculated by dividing the standard deviation (σ) by the square root of the sample size (n):
SE = σ / sqrt(n)
In this case, the sample size (n) is 16 and the standard deviation (σ) is $25.
SE = 25 / sqrt(16)
SE = 25 / 4
SE = 6.25
Next, we need to find the z-scores corresponding to $10 above the mean and $10 below the mean. The z-score is a measure of how many standard errors a particular value is away from the mean. We can calculate the z-scores using the formula:
z = (x - μ) / σ
For $10 above the mean:
z1 = ($100 + $10 - $100) / $25 = 0.4
For $10 below the mean:
z2 = ($100 - $10 - $100) / $25 = -0.4
Now, we can use a standard normal distribution table or calculator to find the cumulative probability associated with these z-scores:
P( -0.4 < z < 0.4)
Using the standard normal distribution table or calculator, we find that the Z-score for 0.4 is 0.6554 and the Z-score for -0.4 is 0.3446. Therefore,
P( -0.4 < z < 0.4) = 0.6554 - 0.3446 = 0.3108
So, the probability that the average cost of repairs will be within $10 of the mean is approximately 0.3108 or 31.08%.