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BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 1 of 16
Unit 7 – The Normal Distribution
Homework
SOLUTIONS
1. This exercise gives you practice in calculating probabilities under the standard normal
curve. See lecture notes for unit 7 page 15. A good url to use is
http://davidmlane.com/hyperstat/z_table.html
Recall the convention of using the letter Z to represent a random variable that is distributed
standard normal. Find the proportion of observations from a standard normal distribution that
satisfies each of the following statements.
a.
b.
c.
d.
e.
f.
g.
h.
Z < 2.85
Z > 2.85
Z > -1.66
-1.66 < Z < 2.85
Z < -2.25
Z > -2.25
Z > 1.77
-2.25 < Z < 1.77
Notes – I used http://davidmlane.com/hyperstat/z_table.html
#1a. Pr (Z < 2.85) = .9978
sol_normal.docx
BIOSTATS 540 – Fall 2016
#1b. Pr (Z > 2.85) = .0022
#1c. Pr (Z > -1.66) = .9515
sol_normal.docx
Introductory Biostatistics
Page 2 of 16
BIOSTATS 540 – Fall 2016
#1d. Pr (-1.66 < Z < 2.85) = .9494
#1e. Pr (Z < -2.25) = .0122
sol_normal.docx
Introductory Biostatistics
Page 3 of 16
BIOSTATS 540 – Fall 2016
#1f. Pr (Z > -2.25) = .9878
#1g. Pr (Z > 1.77) = .0384
sol_normal.docx
Introductory Biostatistics
Page 4 of 16
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 5 of 16
#1h. Pr (-2.25 < Z < 1.77) = .9494
2. This exercise gives you practice in calculating probabilities under normal curves with nonzero mean and non-unit variance. The same url will work for this exercise too.
http://davidmlane.com/hyperstat/z_table.html
The height, X, of young American women is distributed normal with mean µ=65.5 and standard
deviation σ=2.5 inches. Find the probability of each of the following events.
a. X < 67
b. 64 < X < 67
#2a. .7257
⎡⎛ X-µ ⎞ ⎛ 67-µ ⎞ ⎤
pr(X < 67) = pr ⎢⎜
⎟<⎜
⎟⎥
σ
⎠ ⎝ σ ⎠⎦
⎣⎝
⎡
⎛ 67-65.5 ⎞ ⎤
⎟⎥
⎝ 2.5 ⎠ ⎦
. = pr ⎢ Z < ⎜
⎣
= pr [ Z < .6]
= .7257
sol_normal.docx
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
#2b. .4515
⎡⎛ 64-65.5 ⎞
⎛ 67-65.5 ⎞ ⎤
pr ( 64 < X < 67 ) = pr ⎢⎜
⎟ <Z< ⎜
⎟⎥
⎝ 2.5 ⎠ ⎦
⎣⎝ 2.5 ⎠
. = pr [-0.6 < Z < +0.6]
=.4515
sol_normal.docx
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BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 7 of 16
3. This exercise gives you additional practice in calculating probabilities under normal
curves with non-zero mean and non-unit variance.
Suppose that, in a certain population, the distribution of GRE scores is normal with mean µ=600
and standard deviation σ=80.
a. What is the probability of a score less than 450 or greater than 750?
Answer: .0608
Solution:
Define the random variable X = GRE score.
Thus, X is distributed normal with mean µ=600 and standard deviation σ=80.
We write this more compactly as X ~ Normal (µ=600, σ=80). à
Probability { score < 450 OR score > 750 }
= pr [ X < 450] + pr [ X > 750]
= .0608
Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1)
Click on button for “area from value” (2) OUTSIDE (3) values 450 and 750 (4) mean = 600 and (5)
SD=80.
sol_normal.docx
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 8 of 16
b. What proportion of students has scores between 450 and 750?
Answer: .9392
Solution: “Proportion” of students with scores between 450 and 750 à we want:
= pr [ 450 < X < 750]
=.9392
Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1)
Click on button for “area from value” (2) BETWEEN (3) values 450 and 750 (4) mean = 600 and (5)
SD=80.
sol_normal.docx
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 9 of 16
c. What score is equal to the 95th percentile?
Answer: 731.2
There are at least two solutions to this question:
Solution I – Simple “plug in” variety
Solution II – 2 step solution that re-enforces the concepts..
Step 1: Obtain the 95th percentile for Z∼Normal(0,1). Call this Z.95
Step 2: Use Z.95 and the formula on page 19 of the course notes to obtain X.95
Solution I:
Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1)
Click on button for ‘value from an area” (2) area = .95 (3) mean = 600 and (4) SD=80. Don’t forget to
click RECALCULATE.
sol_normal.docx
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 10 of 16
Solution II Step 1:
Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1)
Click on button for “value from an area” (2) BELOW, (3) area = .95 (4) mean = 0 and (5) SD=1.
Solution II Step 2:
Use the formula on page 19 of the unit 5 notes with the following inputs: (1) Z.95 = 1.645 (2) µ = 600
and σ = 80
X .95 = σZ.95 + µ
= ( 80 )[1.645 ] + 600
= 731.6
sol_normal.docx
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 11 of 16
4. The Chapin Social Insight Test evaluates how accurately the subject appraises other
people. In the reference population used to develop the test, Chapin Social Insight Test scores
are distributed normal with mean µ=25 and standard deviation σ=5.
a. What proportion of the population has scores below 20 on the Chapin test?
Answer: .1587
Solution:
The solution for the “proportion of the population” is a probability calculation.
Define the random variable X = Chapin Social Insight Test Score.
X is distributed Normal (µ=25, σ=5).
Want: pr(X < 20) = .1587
Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1)
Click on button for “area from a value” (2) BELOW (3) value =20 (4) mean = 25 and (4) SD=5.
sol_normal.docx
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 12 of 16
b. What proportion has scores below 10?
Answer: .0014
Solution:
This is similar to “a”. à
The solution for the “proportion of the population” is a probability calculation.
X is distributed Normal (µ=25, σ=5).
Want: pr(X < 10) = .0014
url used: http://davidmlane.com/hyperstat/z_table.html
sol_normal.docx
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 13 of 16
c. How high a score must you have in order to be in the top quarter of the population in
social insight?
Answer: 28.35
Solution:
Use the calculator at http://davidlane.com/hyperstat/z_table.html with the following selections: (1)
Click on button for ‘value from an area” (2) ABOVE (3) area = .25 (4) mean = 25 and (5) SD=5.
sol_normal.docx
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 14 of 16
5. A normal distribution has mean µ=100 and standard deviation σ=15 (for example, IQ).
Give limits, symmetric about the mean, within which 95% of the population would lie:
Solution:
This exercise is asking you to work with the following characteristic of the Normal distribution:
If X1, X2, …. Xn are a simple random sample, each distributed Normal(µ, σ2)
Then the sample mean of n observations is distributed Normal((µ, σ2/n)
Tip!
The “David Lane” calculator does not have a special box labeled SE
So, to use the “David Lane” calculator for the normal distribution of a sample mean,
it is necessary to input the value of σ 2 / n in the box “SD”
a) Individual observations
Answer: 70.6, 129.4
“Individual observations” à want to set the “David Lane” boxes as follows:
“MEAN” = µ=100
“SD” = σ = 15.
sol_normal.docx
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
b) Means of 4 observations
Answer: 85.3, 114.7
“Means of 4 observations” à want to set the “David Lane” boxes as follows:
“MEAN” = µ=100
“SD” = SE = √(σ 2/n) = σ/√4 = 15/2 = 7.5
c) Means of 16 observations
Answer: 92.65, 107.35
“Means of 16 observations” à want to set the “David Lane” boxes as follows:
“MEAN” = µ=100
“SD” = SE = √(σ 2/n) = σ/√16 = 15/4 = 3.75
sol_normal.docx
Page 15 of 16
BIOSTATS 540 – Fall 2016
Introductory Biostatistics
Page 16 of 16
d) Means of 100 observations
Answer: 97.06, 102.94
“Means of 100 observations” à want to set the “David Lane” boxes as follows:
“MEAN” = µ=100
“SD” = SE = √(σ 2/n) = σ/√100 = 15/10 = 1.5
sol_normal.docx