Specifications for a part for a DVD player state that the part should weigh between 24 and 25 ounces. The process that produces the parts has a mean of 24.5 ounces and a standard deviation of .2 ounce. The distribution of output is normal.
a. What percentage of parts will not meet the weight specs?
b. Within what values will 95.44 percent of sample means of this process fall, if samples of n = 16 are taken and the process is in control (random)?
The Correct Answer and Explanation is :
Let’s break down the problem step by step:
Part a: Percentage of parts that will not meet the weight specifications
Given:
- Mean weight = 24.5 ounces
- Standard deviation = 0.2 ounces
- Weight specification range = 24 to 25 ounces
To find the percentage of parts that do not meet the weight specifications, we need to calculate the percentage of parts that fall outside the range of 24 to 25 ounces. Since the distribution of output is normal, we can use Z-scores to calculate this probability.
- Calculate the Z-scores for the lower and upper bounds of the specification range: The Z-score formula is: [
Z = \frac{X – \mu}{\sigma}
] Where:
- (X) is the value we are comparing (either 24 or 25 ounces),
- (\mu) is the mean (24.5 ounces),
- (\sigma) is the standard deviation (0.2 ounces). For the lower bound (24 ounces):
[
Z_1 = \frac{24 – 24.5}{0.2} = \frac{-0.5}{0.2} = -2.5
] For the upper bound (25 ounces):
[
Z_2 = \frac{25 – 24.5}{0.2} = \frac{0.5}{0.2} = 2.5
]
- Find the probabilities corresponding to these Z-scores: Using the standard normal distribution table or a calculator, we can find the probabilities associated with Z-scores of -2.5 and 2.5.
- For (Z_1 = -2.5), the probability is approximately 0.0062.
- For (Z_2 = 2.5), the probability is approximately 0.9938.
- Calculate the percentage of parts within the range: The probability of a part falling within the 24 to 25-ounce range is the difference between the two probabilities:
[
P(\text{within range}) = P(Z_2) – P(Z_1) = 0.9938 – 0.0062 = 0.9876
] So, 98.76% of the parts will meet the specifications. - Calculate the percentage that will not meet the specifications: The percentage of parts that fall outside the specification range is:
[
P(\text{outside range}) = 1 – P(\text{within range}) = 1 – 0.9876 = 0.0124
] Therefore, 1.24% of the parts will not meet the weight specifications.
Part b: Range within which 95.44% of sample means will fall (for n = 16)
For sample means, the standard deviation is adjusted to the standard error of the mean. The standard error (SE) is given by:
[
SE = \frac{\sigma}{\sqrt{n}}
]
Where:
- (\sigma = 0.2) ounces,
- (n = 16) (sample size).
[
SE = \frac{0.2}{\sqrt{16}} = \frac{0.2}{4} = 0.05 \, \text{ounces}
]
For a normal distribution, 95.44% of sample means fall within 2 standard deviations of the mean. So, the range of sample means will be:
[
\mu – 2 \times SE \quad \text{to} \quad \mu + 2 \times SE
]
Where (\mu = 24.5) ounces.
[
24.5 – 2 \times 0.05 = 24.4 \, \text{ounces}
]
[
24.5 + 2 \times 0.05 = 24.6 \, \text{ounces}
]
Thus, 95.44% of sample means will fall between 24.4 and 24.6 ounces.
Explanation:
- Part a: We used the normal distribution to calculate the percentage of parts that fall outside the specified weight range (24 to 25 ounces). By finding the Z-scores for both bounds and calculating the probability of falling within the range, we concluded that 1.24% of parts would not meet the weight specifications.
- Part b: For the sample means, we calculated the standard error (SE), which is the standard deviation of the sample mean distribution. Given that 95.44% of data points fall within two standard deviations of the mean in a normal distribution, we used the SE to determine the range within which 95.44% of the sample means will fall, which is between 24.4 and 24.6 ounces.
This explanation shows the application of the Z-score formula and concepts like the standard error for understanding the variation in individual parts versus sample means.