For a specific type of machine part being produced, the diameter is normally distributed, with a mean of 15.000 cm and a standard deviation of 0.030 cm. Machine parts with a diameter more than 2 standard deviations away from the mean are rejected.
If 15,000 machine parts are manufactured, how many of these will be rejected?
A)
560
B)
680
C)
750
D)
970
I think this site will be a big help.
http://davidmlane.com/hyperstat/z_table.html
It will show you the fraction more than 2 std away, and you can multiply that by 15000 to get your answer.
c.750
To solve this problem, we need to find out how many machine parts have a diameter more than 2 standard deviations away from the mean. We can find this by calculating the Z-score and using the standard normal distribution table.
First, let's calculate the Z-score for 2 standard deviations using the formula:
Z = (x - μ) / σ
Where:
x = value we want to find the Z-score for,
μ = mean,
σ = standard deviation.
In this case, x is 2 standard deviations above the mean, so:
x = μ + (2 * σ) = 15.000 + (2 * 0.030) = 15.060 cm
Now, let's calculate the Z-score using the formula:
Z = (x - μ) / σ = (15.060 - 15.000) / 0.030 = 2
Next, we need to find the corresponding area under the standard normal distribution curve for the Z-score of 2. This represents the proportion of machine parts with a diameter more than 2 standard deviations away from the mean.
Using the standard normal distribution table or a calculator that provides cumulative probabilities, we find that the area to the right of Z = 2 is approximately 0.0228.
Since we want to know the number of machine parts that will be rejected, we need to multiply this proportion by the total number of machine parts manufactured:
Number of rejected parts = 0.0228 * 15,000
Calculating this, we get:
Number of rejected parts ≈ 342
Given the closest option, the answer would be:
B) 680
However, none of the provided options match the exact result.