To solve this problem, we will use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean. Here are the steps: 1. Identify the given values: - Population mean (μ) = 89 - Population standard deviation (σ) = 5 - Sample size (n) = 123 - Sample mean (x̄) = 87.8 2. Calculate the standard error of the mean (SE). The standard error of the mean is the standard deviation of the sampling distribution of the mean. It is calculated as: SE = σ / √n SE = 5 / √123 SE ≈ 0.45 3. Calculate the Z-score. The Z-score is calculated as: Z = (x̄ - μ) / SE Z = (87.8 - 89) / 0.45 Z ≈ -2.67 4. Look up the Z-score in the Z-table to find the probability. The Z-table tells us what percentage of the population is below (to the left of) our score. Since we want to find the probability that the mean monitor life is greater than 87.8 months, we need to find the area to the right of our Z-score. The Z-table gives us the area to the left, so we subtract the value from 1 (the total area under the curve) to get the area to the right: P(Z -2.67) = 1 - P(Z -2.67) = 1 - 0.0038 P(Z -2.67) = 0.9962 So, if the production manager's claim is true, the probability that the mean monitor life would be greater than 87.8 months in a sample of 123 monitors is approximately 0.9962 or 99.62% when rounded to four decimal places.

Question

To solve this problem, we will use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean.

Here are the steps:

1. Identify the given values:
   - Population mean (μ) = 89
   - Population standard deviation (σ) = 5
   - Sample size (n) = 123
   - Sample mean (x̄) = 87.8

2. Calculate the standard error of the mean (SE). The standard error of the mean is the standard deviation of the sampling distribution of the mean. It is calculated as:

SE = σ / √n
   SE = 5 / √123
   SE ≈ 0.45

3. Calculate the Z-score. The Z-score is calculated as:

Z = (x̄ - μ) / SE
   Z = (87.8 - 89) / 0.45
   Z ≈ -2.67

4. Look up the Z-score in the Z-table to find the probability. The Z-table tells us what percentage of the population is below (to the left of) our score. Since we want to find the probability that the mean monitor life is greater than 87.8 months, we need to find the area to the right of our Z-score.

The Z-table gives us the area to the left, so we subtract the value from 1 (the total area under the curve) to get the area to the right:

P(Z > -2.67) = 1 - P(Z < -2.67)
   P(Z > -2.67) = 1 - 0.0038
   P(Z > -2.67) = 0.9962

So, if the production manager's claim is true, the probability that the mean monitor life would be greater than 87.8 months in a sample of 123 monitors is approximately 0.9962 or 99.62% when rounded to four decimal places.

Knowee AI · Accepted Answer