Answer :
The probability that the average number of days where breakfast got skipped is between 48 and 51 times for a random sample of 36 Americans is approximately 0.9311, or 93.11%.
To find the probability that the average number of days where breakfast got skipped is between 48 and 51 times for a random sample of 36 Americans, we'll use the Central Limit Theorem.
The Central Limit Theorem states that the distribution of the sample means will be approximately normally distributed, regardless of the shape of the original population, as long as the sample size is sufficiently large (typically n ≥ 30).
Given that the population mean (μ) is 50 times and the population standard deviation (σ) is 4 days, we can calculate the standard deviation of the sample mean (standard error) using the formula:
Standard Error (SE) = σ / √n,
where n is the sample size (36 in this case).
SE = 4 / √36 = 4 / 6 = 2/3 ≈ 0.67.
Now, we need to find the z-scores for the sample means of 48 and 51:
Z-score for 48: (48 - μ) / SE = (48 - 50) / 0.67 ≈ -2.99.
Z-score for 51: (51 - μ) / SE = (51 - 50) / 0.67 ≈ 1.49.
Next, we find the probabilities associated with these z-scores using a standard normal distribution table or a calculator:
P(Z ≤ -2.99) ≈ 0.0014 (probability for 48 or less times).
P(Z ≤ 1.49) ≈ 0.9325 (probability for 51 or less times).
To find the probability between 48 and 51 times, we subtract the probability for 48 or less times from the probability for 51 or less times:
P(48 ≤ X ≤ 51) ≈ P(Z ≤ 1.49) - P(Z ≤ -2.99) ≈ 0.9325 - 0.0014 ≈ 0.9311.
Therefore, the probability that the average number of days where breakfast got skipped is between 48 and 51 times for a random sample of 36 Americans is approximately 0.9311, or 93.11%.
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