For a normal distribution with µ = 200 and ó = 50, what X values form the boundaries for the middle 60% of the distribution?
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion (± .30 to mean) to find the Z score. Insert into equation and solve. Remember to double to represent each side.
242.081, 157.919
To find the boundaries for the middle 60% of the distribution, you'll need to use the z-score formula and the standard normal distribution table.
Step 1: Convert the given mean (µ) and standard deviation (ó) to z-scores.
In this case, the mean (µ) is given as 200 and the standard deviation (ó) is given as 50.
Formula for z-score: z = (X - µ) / ó
For the lower boundary, to find the z-score corresponding to the X value separating the lower 20% of the distribution, you would use the formula:
lower_boundary_z = invNorm(0.20)
where "invNorm" is the inverse of the normal distribution function.
For the upper boundary, to find the z-score corresponding to the X value separating the upper 20% of the distribution, you would use the formula:
upper_boundary_z = invNorm(0.80)
Step 2: Convert the z-scores back to X values.
To convert the z-scores back to X values, we use the formula:
X = z * ó + µ
Step 3: Calculate the lower and upper boundaries for the middle 60% of the distribution.
Using the X = z * ó + µ formula, you can calculate the lower and upper boundaries for the middle 60% of the distribution.
Lower boundary: X_lower = lower_boundary_z * 50 + 200
Upper boundary: X_upper = upper_boundary_z * 50 + 200
By plugging in the values, you can calculate the X values for the boundaries of the middle 60% of the distribution.