Suppose the weights of Farmer Carl's potatoes are normally distributed with a mean of 7.9 ounces and a standard deviation of 1.1 ounces.
Suppose Carl bags his potatoes in randomly selected groups of 6. What percentage of these bags should have a mean potato weight between 7.5 and 8.5 ounces?
If each potato is 7.9 ± 1.1, then none of the BAGS of 6 will be between 7.5 and 8.5. For individual potatoes:
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability between the two Z scores. Multiply by 100.
To find the percentage of bags that should have a mean potato weight between 7.5 and 8.5 ounces, we can use the Central Limit Theorem.
The Central Limit Theorem states that, regardless of the shape of the population distribution, the distribution of the sample means will approach a normal distribution as the sample size increases.
In this case, we have a normal distribution with a mean (μ) of 7.9 ounces and a standard deviation (σ) of 1.1 ounces. Since Carl bags his potatoes in groups of 6, the sample size (n) is 6.
To calculate the mean and standard deviation of the sample means, we divide the population mean by the square root of the sample size and divide the population standard deviation by the square root of the sample size:
Sample Mean (μ') = μ / √n --> μ' = 7.9 / √6 --> μ' ≈ 3.214
Sample Standard Deviation (σ') = σ / √n --> σ' = 1.1 / √6 --> σ' ≈ 0.449
Now, we need to find the z-scores corresponding to the sample mean values of 7.5 and 8.5 ounces using the z-score formula:
z = (x - μ') / σ'
where x is the sample mean.
z1 = (7.5 - 3.214) / 0.449 ≈ 9.70
z2 = (8.5 - 3.214) / 0.449 ≈ 12.66
Now, we can use a standard normal distribution table or a calculator to find the cumulative probability (percentage) between these z-scores.
Typically, these types of tables provide the cumulative probability to the left of the given z-score. Since we want the probability between these two z-scores, we subtract the cumulative probability corresponding to the lower z-score from the cumulative probability corresponding to the higher z-score.
P(7.5 ≤ x ≤ 8.5) = P(z1 ≤ Z ≤ z2) = P(Z ≤ z2) - P(Z ≤ z1)
Using a standard normal distribution table or calculator, we find:
P(Z ≤ 12.66) ≈ 1.0 (since it is a very large value, practically close to 1)
P(Z ≤ 9.70) ≈ 1.0
Therefore,
P(7.5 ≤ x ≤ 8.5) ≈ 1.0 - 1.0 = 0
Hence, approximately 0% of the bags should have a mean potato weight between 7.5 and 8.5 ounces.