Suppose that the average amount paid by customers at a grocery store is $80. The standard deviation of the amounts paid is $24. The distribution of the amounts paid is skewed. (Notice that the individuals here are customers and the variable is the amount paid per customer.) Consider randomly selecting 40 customers and calculating the average amount paid per customer in the sample. Do this many times. What percentage of all the sample averages would exceed $75? (That is, what is the chance that the average of 40 randomly selected amounts paid exceeds $75?) Please show all the steps as well as an appropriate diagram
To calculate the percentage of sample averages that would exceed $75, we need to use the Central Limit Theorem and assume that the sample averages follow a normal distribution.
Step 1: Find the standard deviation of the sample means.
According to the Central Limit Theorem, the standard deviation of the sample means (also known as the standard error) can be calculated using the formula:
σ(sample mean) = σ(population) / √(n)
where σ(population) is the standard deviation of the population, and n is the sample size.
In this case, σ(population) = $24, and n = 40.
So, σ(sample mean) = 24 / √(40) ≈ $3.79
Step 2: Calculate the z-score of the threshold value.
To determine the chance that the average of 40 randomly selected amounts paid exceeds $75, we need to convert this value into a z-score.
The formula to calculate the z-score is:
z = (x - μ) / σ
where x is the threshold value, μ is the population mean, and σ is the standard deviation.
In this case, x = $75, μ = $80, and σ(sample mean) = $3.79.
z = (75 - 80) / 3.79 ≈ -1.32
Step 3: Find the percentage using the z-score table.
Look up the z-score of -1.32 in the z-score table (or use a calculator/online tool). The z-score table shows that the cumulative probability corresponding to -1.32 is approximately 0.0934.
However, we are interested in the probability of exceeding $75, so we subtract the cumulative probability from 1:
P(X > 75) = 1 - 0.0934 ≈ 0.9066
Step 4: Convert the probability to percentage.
To find the percentage, multiply the probability by 100:
Percentage = 0.9066 * 100 ≈ 90.66%
Therefore, approximately 90.66% of all the sample averages would exceed $75.
As for the diagram, you can visually represent the normal distribution and shade the area to the right of $75.
To find the percentage of all the sample averages that would exceed $75, we need to calculate the probability using the Central Limit Theorem. The Central Limit Theorem states that if we have a sufficiently large sample size, the distribution of sample means will approach a normal distribution regardless of the shape of the population distribution.
Here are the steps to calculate the probability:
Step 1: Determine the population mean and standard deviation.
- Population mean (μ) = $80
- Population standard deviation (σ) = $24
Step 2: Determine the sample size.
- Sample size (n) = 40
Step 3: Find the standard error of the mean (σᵦ).
- Standard error of the mean (σᵦ) = σ / √n
- σᵦ = $24 / √40
Step 4: Calculate the z-score.
- z = (X - μ) / σᵦ
- X = $75
- z = ($75 - $80) / ($24 / √40)
Step 5: Find the probability using the z-table.
- Look up the z-score from step 4 in the z-table to find the corresponding probability. The probability will be the area under the normal distribution curve to the right of the z-score.
Now, I'll calculate the probability using these steps:
Step 1: μ = $80, σ = $24
Step 2: n = 40
Step 3: σᵦ = $24 / √40 ≈ $3.79
Step 4: z = ($75 - $80) / ($3.79) ≈ -1.32
Step 5: Using the z-table, the probability of z > -1.32 is approximately 0.9066.
So, there is about a 90.66% chance that the average of 40 randomly selected amounts paid exceeds $75.
Here is an appropriate diagram to illustrate the calculation:
```
___
\ \
____________\ \_________
\ / \ /
\________/______\_____/
-1.32 0
```
The shaded area to the right of the z-score (-1.32) represents the probability that the sample average exceeds $75.