Answer:
Explanation:
To find three standard deviations from the mean, we can multiply the standard deviation by 3 and add/subtract the result from the mean:
Three standard deviations above the mean:
27 + (3 * 3) = 27 + 9 = 36
Three standard deviations below the mean:
27 - (3 * 3) = 27 - 9 = 18
So, the labels are:
Mean (μ) = 27
Three standard deviations above the mean = 36
Three standard deviations below the mean = 18
b) To find the percentage of scores between 24 and 30, we need to calculate the area under the normal distribution curve within that range. Since the distribution is assumed to be normal, we can use the properties of the normal distribution.
Using a standard normal distribution table or a calculator, we can find the area to the left of 30 (Z-score for 30) and subtract the area to the left of 24 (Z-score for 24). The difference gives us the percentage of scores between 24 and 30.
Let's calculate the Z-scores for 24 and 30:
Z-score for 24 = (24 - μ) / σ = (24 - 27) / 3 = -1
Z-score for 30 = (30 - μ) / σ = (30 - 27) / 3 = 1
From the standard normal distribution table or using a calculator, we find:
Area to the left of Z = -1 is approximately 0.1587.
Area to the left of Z = 1 is approximately 0.8413.
Percentage of scores between 24 and 30 = 0.8413 - 0.1587 = 0.6826 = 68.26%
Therefore, approximately 68.26% of scores are between 24 and 30.
c) To approximate the number of juniors who scored between 27 and 30, we can assume that the distribution is continuous and use the percentage calculated in part (b).
Percentage of scores between 27 and 30 = 68.26%
Total number of juniors who took the ACT = 250
Approximately, the number of juniors who scored between 27 and 30 is:
Number of juniors = Percentage * Total number of juniors
Number of juniors = 0.6826 * 250
Number of juniors ≈ 170
Therefore, approximately 170 juniors scored between 27 and 30.