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 the answer is B
To find the number of machine parts that will be rejected, we need to determine the area under the normal distribution curve that represents the probability of getting a diameter more than 2 standard deviations away from the mean.
First, we calculate the value of 2 standard deviations away from the mean:
2 * 0.030 cm = 0.060 cm
Next, we find the z-score for this value using the formula:
Z = (X - μ) / σ
where X is the value, μ is the mean, and σ is the standard deviation.
Z = (0.060 - 15.000) / 0.030
Z = -14.940 / 0.030
Z ≈ -498
Looking up the value -498 in the standard normal distribution table, we find that the area to the left of this value is approximately 0. This means that the probability of getting a diameter more than 2 standard deviations away from the mean is very close to 0.
Now, we can calculate the number of rejected machine parts by finding the probability of a diameter being more than 2 standard deviations away using the normal distribution equation and then multiplying it by the total number of machine parts:
Number of rejected machine parts = Probability of rejection * Total number of machine parts
Number of rejected machine parts = (1 - Probability of a diameter being within 2 standard deviations away) * 15000
Number of rejected machine parts = (1 - 0) * 15000
Number of rejected machine parts = 0 * 15000
Number of rejected machine parts = 0
Therefore, the correct answer is none of the given options (0 machine parts will be rejected).
To determine the number of machine parts that will be rejected, we need to calculate the z-score for the cutoff point, which is 2 standard deviations away from the mean.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
x = Value of the variable (cutoff point)
μ = Mean of the distribution
σ = Standard deviation of the distribution
In this case, the cutoff point is 2 standard deviations away from the mean, so the z-score will be 2.
Using the given values:
μ = 15.000 cm
σ = 0.030 cm
z = 2
Now, we can calculate the value of x by rearranging the formula:
x = z * σ + μ
Substituting the values:
x = 2 * 0.030 + 15.000
x = 0.060 + 15.000
x = 15.060 cm
Therefore, any machine parts with a diameter greater than 15.060 cm will be rejected.
To find the number of rejected machine parts, we need to find the proportion of the distribution that falls to the right of 15.060 cm.
Using a standard normal distribution table, we can find the area to the left of 15.060 cm. In the table, this corresponds to the z-score of 2.
The table value for a z-score of 2 is approximately 0.9772.
To find the proportion to the right of 15.060 cm, we subtract the table value from 1:
amount rejected = 1 - 0.9772
amount rejected = 0.0228
Finally, we multiply the proportion by the total number of machine parts to find the number of rejected machine parts:
number of rejected machine parts = 0.0228 * 15000
number of rejected machine parts ≈ 342
Therefore, the correct answer is not among the options provided.