Question

A normal distribution has a mean of m = 60 and a

standard deviation of s = 16. For each of the following

scores, indicate whether the body is to the right or

left of the score and find the proportion of the distribution

located in the body.

a. X = 64

b. X = 80

c. X = 52

d. X = 28

Answer 1

Solution:

A normal distribution has a
Mean (m)= 60
Standard deviation (s) = 16
We need to find that score indicate whether the body is to the right or left of the score and proportion of the distribution as follows:
For this, we will found Z score as follows
Solution(a)
For X= 64
Z-Score = (X -m)/s = (64-60)/16 = 0.25
So this score 64 is to the right of the score or mean score and the percentile of the score from the Z table is 0.5987 or 59.87%
Solution(b)
For X= 80
Z-Score = (X -m)/s = (80-60)/16 = 1.25
So this score 80 is to the right of the score or mean score and the percentile of the score from the Z table is 0.8944 or 89.44%
Solution(c)
For X= 52
Z-Score = (X -m)/s = (52-60)/16 = -0.5
So this score 52 is to the left of the score or mean score and the percentile of the score from the Z table is 0.3085 or 30.85%
Solution(d)
For X= 28
Z-Score = (X -m)/s = (28-60)/16 = -2
So this score 28 is to the left of the score or mean score and the percentile of the score from the Z table is 0.0228 or 2.28%