The diameter of a pipe is normally distributed with a mean of 0.4 inches and a variance of 0.0004. What is the probability that the diameter of a randomly selected pipe will exceed 0.44 inches?
If no one who teaches this shows up to help, I will get out my normal distribution tables and try later.
As Damon suggested, these are done using Normal distribution tables.
Personally I don't see the difference in using tables or computer programs that do the same thing.
My personal choice is
http://davidmlane.com/hyperstat/z_table.html
If I recall correctly, there is a difference in "variance" and "standard deviation"
check in your textbook
the standard deviation is the square root of the variance.
To find the probability that the diameter of a randomly selected pipe will exceed 0.44 inches, we need to calculate the area under the normal distribution curve to the right of 0.44.
First, we need to calculate the standard deviation. The variance (σ^2) is given as 0.0004, so the standard deviation (σ) is the square root of the variance, which is √(0.0004) = 0.02 inches.
Next, we need to standardize the value 0.44 using the following formula:
Z = (X - μ) / σ
where X is the value we want to standardize (0.44), μ is the mean (0.4), and σ is the standard deviation (0.02).
Plugging in the values, we get:
Z = (0.44 - 0.4) / 0.02 = 2
Now, we need to find the probability corresponding to the standardized value of 2. We can consult a standard normal distribution table or use a calculator to find this probability. Let's assume we use a standard normal distribution table.
Looking up the value of 2 in the standard normal distribution table, we find that the area to the left of 2 is 0.9772. Since we want the probability to the right of 0.44, we subtract this value from 1.
1 - 0.9772 = 0.0228
Therefore, the probability that the diameter of a randomly selected pipe will exceed 0.44 inches is approximately 0.0228, or 2.28%.