A distribution of values is normal with a mean of 130 and a standard deviation of 5.
Find the interval containing the middle-most 38% of scores:
To find the interval containing the middle-most 38% of scores in a normal distribution, we can use the properties of the normal distribution and the concept of Z-scores. The middle-most 38% of scores will be centered around the mean and will span from the 31st percentile to the 69th percentile because 50% - (38%/2) = 31% and 50% + (38%/2) = 69% (since the distribution is symmetric around the mean).
To find these percentile boundaries, we will use a standard normal distribution table or a calculator that provides the Z-scores corresponding to given percentiles.
The Z-score associated with the 31st percentile (0.31 in the standard normal distribution) is approximately -0.50 and the Z-score associated with the 69th percentile (0.69 in the standard normal distribution) is approximately 0.50. These values might slightly differ depending on the Z-score table or the precision of the tool being used, but they should be close to these approximate values.
Given the mean (μ) and standard deviation (σ) of the normal distribution, we can compute the actual scores corresponding to these Z-scores using the formula:
X = μ + Z * σ
For the 31st percentile (lower boundary):
X_low = 130 + (-0.50) * 5
X_low = 130 - 2.5
X_low = 127.5
For the 69th percentile (upper boundary):
X_high = 130 + (0.50) * 5
X_high = 130 + 2.5
X_high = 132.5
Therefore, the interval containing the middle-most 38% of the scores in a normal distribution with a mean of 130 and a standard deviation of 5 is approximately from 127.5 to 132.5.