A manufacturing process produces a precision part whose length is a normal random variable with mean 80 millimeters and standard deviation 2 millimeters. All parts deviating by more than 4.8 millimeters from the mean must be rejected. On average, what percentage of the parts must be rejected?
The percentage of parts that must be rejected is $\square \%$.
(Type an integer or decimal rounded to two decimal places as needed.)
Answer
Final Answer: The percentage of parts that must be rejected is \(\boxed{1.64\%}\).
Step 1 ：We are given a manufacturing process that produces a precision part whose length is a normal random variable with a mean of 80 millimeters and a standard deviation of 2 millimeters. All parts deviating by more than 4.8 millimeters from the mean must be rejected. We are asked to find the average percentage of the parts that must be rejected.
Step 2 ：First, we need to calculate the number of standard deviations that 4.8 millimeters represents. Since the standard deviation is 2 millimeters, 4.8 millimeters is \(\frac{4.8}{2} = 2.4\) standard deviations from the mean.
Step 3 ：Next, we calculate the probability of a part being within 2.4 standard deviations of the mean. This can be done using the cumulative distribution function (CDF) of the normal distribution. The CDF at 2.4 is approximately 0.9918.
Step 4 ：We then calculate the probability of a part being outside this range, which is 1 minus the probability of a part being within 2.4 standard deviations of the mean. This gives us approximately 0.0082.
Step 5 ：Since the distribution is symmetric, the percentage of parts that must be rejected is twice this value. Therefore, the percentage of parts that must be rejected is \(2 \times 0.0082 \times 100 = 1.64\%\).
Step 6 ：Final Answer: The percentage of parts that must be rejected is \(\boxed{1.64\%}\).
