Please I Need Help With This Question Showing All Steps...Thank You :)
The average number of units that the factory workers at Benzo company assemble per week is μ = 45 with a standard deviation of σ = 12. Assume that the distribution of units assembled is normal. If a sample n = 25 were drawn from all the workers,
A. What range units manufactured by this sample would contain the sample mean 90% of the time? (Hint: Find the middle 90% of the distribution of sample means)
B. What is the probability that the mean will be between 40 and 50 units?
C. Is it reasonable for this sample to produce an average of 48 units per week, or is this mean very different from what would be normally expected?
D. How likely is it that this sample will produce more than an average of 55 units?
Thanks You!
Certainly! Let's go through each question step by step:
A. To find the range of units that would contain the sample mean 90% of the time, we need to determine the range for the central 90% of the distribution.
1. Firstly, we need to calculate the standard deviation of the sample mean. This is also known as the standard error, which is given by σ/√n, where σ is the population standard deviation (12 in this case) and n is the sample size (25 in this case).
Standard error = 12/√25 = 12/5 = 2.4
2. Next, we need to determine the critical z-values that correspond to a central area of 90%. We will use the z-table or a statistical calculator for this.
The area to the left of the first critical z-value would be (1-0.9)/2 = 0.05. In the z-table, a value of 0.05 corresponds to a z-score of -1.645.
The area to the right of the second critical z-value would be (1-0.9)/2 = 0.05. In the z-table, a value of 0.05 corresponds to a z-score of 1.645.
3. Now, we can calculate the range of units that would contain the sample mean with a 90% probability.
Range = Sample mean ± (z-value * standard error)
Range = 45 ± (1.645 * 2.4)
Range = 45 ± 3.956
Range = (41.044, 48.956)
Therefore, the range of units that would contain the sample mean 90% of the time is (41.044, 48.956).
B. To find the probability that the mean will be between 40 and 50 units, we need to calculate the z-scores for both values and then find the area under the curve between those two z-scores.
1. For 40 units: z-score = (40 - 45) / (12 / √25) = -2.0833
2. For 50 units: z-score = (50 - 45) / (12 / √25) = 2.0833
3. Using the z-table or a statistical calculator, find the area to the left of each z-score.
Area for z = -2.0833 = 0.0180
Area for z = 2.0833 = 0.9820
4. The probability that the mean will be between 40 and 50 units is the difference between the two areas.
Probability = 0.9820 - 0.0180 = 0.9640
Therefore, the probability that the mean will be between 40 and 50 units is 0.9640, or 96.4%.
C. To determine if it is reasonable for this sample to produce an average of 48 units per week, we need to assess how different it is from the expected mean.
1. Calculate the z-score for the sample mean of 48 units:
z-score = (48 - 45) / (12 / √25) = 1.25
2. Using the z-table or a statistical calculator, find the area to the left of the z-score.
Area for z = 1.25 = 0.8944
3. The probability associated with a z-score of 1.25 is 0.8944, which means that the sample mean of 48 units falls within the top 10.56% of the distribution.
This suggests that producing an average of 48 units per week is not very different from what would be normally expected, as it falls within a reasonably high percentile of the distribution.
D. To determine the likelihood of the sample producing more than an average of 55 units, we need to calculate the probability associated with that value:
1. Calculate the z-score for the value of 55 units:
z-score = (55 - 45) / (12 / √25) = 2.0833
2. Using the z-table or a statistical calculator, find the area to the right of the z-score.
Area for z > 2.0833 = 1 - Area for z < 2.0833 = 1 - 0.9820 = 0.0180
3. The likelihood that the sample will produce more than an average of 55 units is 0.0180, or 1.8%.
Therefore, there is a 1.8% chance that this sample will produce more than an average of 55 units.
I hope the explanations and calculations were clear! Let me know if you have any further questions.