The diameter of an electric cable is normally distributed, with a mean of 0.8 inch and a standard deviation of 0.03 inch. What is the probability that the diameter will exceed 0.83 inch? (You may need to use the standard normal distribution table. Round your answer to three decimal places.)
Thank you very much for your time.
Tools like this below replace tables traditionally found at the back of textbooks
http://davidmlane.com/hyperstat/z_table.html
enter your mean and sd,
click on "above" for "exceeds 0.83" and enter the data
to get .1587 ,or
.159 correct to 3 decimals
To solve this problem, we will use the standard normal distribution table. Since we know the mean and standard deviation of the diameter (0.8 inch and 0.03 inch respectively), we can convert the given value of 0.83 inch into the corresponding z-score using the formula:
z = (x - μ) / σ
where:
x = given value
μ = mean
σ = standard deviation
Plugging in the values:
z = (0.83 - 0.8) / 0.03
Calculating the z-score:
z = 0.03 / 0.03
z = 1
Now that we have the z-score for the given value, we can find the probability of the diameter exceeding 0.83 inch by looking up the z-score in the standard normal distribution table. The table will give us the corresponding cumulative probability.
From the table, we find that the cumulative probability for a z-score of 1 is approximately 0.8413.
However, we want to find the probability of the diameter exceeding 0.83 inch, which is equivalent to finding the probability of having a z-score greater than 1. Since the standard normal distribution table gives the cumulative probability up to a certain z-score, we need to subtract the cumulative probability obtained (0.8413) from 1 to get the desired probability.
P(Z > 1) = 1 - 0.8413 = 0.1587
Therefore, the probability that the diameter will exceed 0.83 inch is approximately 0.159 (rounded to three decimal places).