# ANSWERS FOR Lab 3, Part II

advertisement ```ANSWERS FOR Lab 3, Part II
1. Assume the variable y (age) is normally distributed with a mean of 45 and a standard deviation
of 15.
a. What proportion of the cases are:
i. Younger than 50?
ii. Older than 58?
iii. Between the ages of 43 and 51?
Z-scores for the distribution of individual cases, Mean = 45; standard deviation = 15
Use the formula for Z scores:
Z
y Y
sy
a. Determining proportions:
i. z = 0.33 ; proportion &gt; .33 = .3707; proportion &lt; 50 = (1-.3707) = .6293
The proportion of cases below the age of 50 is .6293. (In percent, 62.93% are
below the age of 50).
ii.
z = 0.87 ; proportion &gt; 0.87 = .1922; proportion &gt; 58 = .1922
The proportion of cases older than 58 is .1922 (19.22% are older than 58
years of age).
iii. z = -0.13, 0.40 ; proportion &lt;-0.13 = .4483; proportion &gt; 0.40=.3446;
proportion between 43 and 51 = 1 - (.4483 + .3446) = .2071
The proportion of cases between 43 and 51 is .2071 (20.71% are between 43
and 51).
b. Find the age (or ages) of the following:
HINT: First, look up the proportion in the Z-table to get the z-score, and then solve the equation
above for “y” instead of Z.
i. which divides the distribution in half.
Proportion &gt; (Z=0) = .5. y = 45
Forty-five years (45) is the age which divides the distribution in half (this is a
give away question, as the mean is always what divides the distribution in
half, on a normal curve).
ii.
above which 25% of the cases fall.
Proportion &gt; (Z=.67) = .25. y= 55.05 years.
Twenty-five percent of the cases are above 55.05 years of age.
iii.
below which 5% of the cases fall.
Proportion &lt; (Z= -1.64) = .05. y = 20.4
The age below which 5% of the cases fall is 20.4 years.
iv.
above which 2.5% of the cases fall.
Proportion &gt; (Z=1.96) = .025.
Two and a half percent of the cases are older than 74.4 years.
v.
below which 1% of the cases fall.
Proportion &lt; (Z= -2.33) = .01.
One percent of the cases are younger than 10.05 years.
vi. above which 0.5% of the cases fall.
Proportion &gt; (Z=2.58) = .005. One half of one percent (0.5%) of the cases are
older than 83.7 years.
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