A production process is checked perifically by a quality control inspector. The inspector selects simple random samples of 30 finished products and computes the sample mean product weights x. If test results over a long period of time show that 5% of the x values are over 2.1 pounds and 5% are under 1.9 pounds, what are the mean and the standard deviation for the population of products produced with this process?
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To find the mean and standard deviation for the population of products produced with this process, we need to use the information given about the sample mean and the percentages of x values that are above 2.1 pounds and below 1.9 pounds.
Let's start by finding the mean. We know that the sample mean is x, and we can assume that the sample mean is an unbiased estimator of the population mean. Therefore, the sample mean equals the population mean. We can write this as:
Population mean = x
Next, let's find the standard deviation. We know that the sample mean follows a normal distribution. From the given information, we can deduce that:
- 5% of the x values are above 2.1 pounds, which means that the upper tail area is 0.05.
- 5% of the x values are below 1.9 pounds, which means that the lower tail area is 0.05.
Since the sample mean follows a normal distribution, we can use the Z-score to find the corresponding values for the upper and lower tail areas. The formula to calculate the Z-score is:
Z = (x - μ) / (σ / sqrt(n))
where Z is the Z-score, x is the sample mean, μ is the population mean, σ is the standard deviation, and n is the sample size.
For the upper tail area of 0.05, we want to find the Z-score such that P(Z > Z-score) = 0.05. Using a Z-table or a statistical calculator, we can find that the Z-score is approximately 1.645.
For the lower tail area of 0.05, we want to find the Z-score such that P(Z < Z-score) = 0.05. Using the property of symmetry of the normal distribution, we can deduce that the Z-score is also approximately -1.645.
Now, we can set up two equations using the Z-score formula:
1.645 = (2.1 - x) / (σ / sqrt(30))
-1.645 = (1.9 - x) / (σ / sqrt(30))
Solving these equations simultaneously will allow us to find both the mean (x) and the standard deviation (σ) for the population of products produced with this process.
Note: The above explanation assumes that the sample mean and the population mean are the same. This assumption is made under the assumption that the sampling process is unbiased and the sample is representative of the population.