Chapter 8 Review Exercises
Problem 248
Since we are after a “…probability distribution…” we need to see how it is
distributed; so first list the numbers and then note how many there are of
each of these numbers:
N
u
m
b
e
r
H
o
w
M
a
n
y
2
3
4
5
6
7
9
1
6
2
3
1
2
1
Now since a probability distribution requires a probability, which is how many
divided by the total, we add them up for the total (1+6+2+3+1+2+1 = 16) and
then divide each:
N
u
m
b
e
r
P
r
o
b
a
b
i
l
i
t
y
2
3
4
5
6
7
9
1
/
1
6
6
/
1
6
2
/
1
6
3
/
1
6
1
/
1
6
2
/
1
6
1
/
1
6
o
r
0
.
0
6
2
5
0
.
3
7
5
0
0
.
1
2
5
0
0
.
1
8
7
5
0
.
0
6
2
5
0
.
1
2
5
0
0
.
0
6
2
5
Problem 249
Taking the probability distribution from Problem and create a histogram (or
bar graph) of the data (you must include the gap between 7 and 9, for the zero
probability of 8):
Problem 2
0.4000
0.3500
0.3000
0.2500
0.2000
0.1500
0.1000
0.0500
0.0000
2
3
4
5
6
7
8
Series 1
Problem 250
The mean is the same as the average, so there are two ways to accomplish
this: (1) add them up and divide by how many numbers or (2) take the sum of
the number times the probability.
(1) (4+3+3+5+7+2+4+3+5+9+3+6+5+3+7+3)/16 = 72/16 = 4.5
(2) 2*0.0625+3*0.375+4*0.125+5*0.1875+6*0.0625+7*0.125+9*0.0625 =
4.5
Problem 251
The expected value is the same as the average (which is also the same as the
mean…see problem 3), so there are two ways to accomplish this: (1) add them
9
up and divide by how many numbers or (2) take the sum of the number times
the probability.
(1) (4+3+3+5+7+2+4+3+5+9+3+6+5+3+7+3)/16 = 72/16 = 4.5
(2) 2*0.0625+3*0.375+4*0.125+5*0.1875+6*0.0625+7*0.125+9*0.0625 =
4.5
Problem 252
The variance (σ2) may be found in one of two ways: (1) using the definition
formula [∑( (𝑥𝑖 − 𝜇)2 𝑝(𝑥))] or (2) using the computing formula [E(x2) µ2]…the mean (µ) comes from problems 250 or 251.
(1) (2-4.5)2*0.0625 + (3-4.5)2*0.375 + (4-4.5)2*0.125 + (5-4.5)2*0.1875 +
(6-4.5)2*0.0625 + (7-4.5)2*0.125 +
(9-4.5)2*0.0625 = 3.5
(2) First calculate E(x2):
22*0.0625+32*0.375+42*0.125+52*0.1875+62*0.0625+72*0.125+92*
0.0625 = 23.75
So the variance is 23.75 – (4.5)2 = 3.5
Now since standard deviation (σ) is the square root of variance (σ2), it is √3.5 =
1.8708
Problem 253(a)
Since we are given:
X
P(X)
-2
1/12
1
¼
3
?
And we know that the probability must add to ONE, then 1/12 + ¼ + P(3) = 1,
so P(3) = 1 – 1/12 – ¼ = 8/12 = 2/3 or 0.6667
Problem 253(b)
For probabilities greater than -2, we add all probabilities that are greater than
-2, giving us
P(1) + P(3) = ¼ + 2/3 = 11/12 or 0.9167
This can also be done using the complement rule, since the only probability
not included is the P(-2), giving us 1 – P(-2) =1 – 1/12 = 11/12 or 0.9167
Problem 253(c)
For probabilities greater than or equal to -2, we add all probabilities that are
greater than -2 as well as the probability of -2, which means you have
everything (all the probability): 1
Problem 253(d)
The expected value is the sum of the number times the probability:
-2*1/12+1*1/4+3*2/3 = -2/12 + 3/12 + 2 = 25/12 or 2.0833
Problem 253(e)
The variance (σ2) may be found in one of two ways: (1) using the definition
formula [∑( (𝑥𝑖 − 𝜇)2 𝑝(𝑥))] or (2) using the computing formula [E(x2) µ2]…the mean (µ) comes from problem 256.
(1) (-2-25/12)2*1/12 + (1-25/12)2*1/4 + (3-25/12)2*2/3 = 2.2431 or
323/144
(2) First calculate E(x2): -22*1/12+12*1/4+32*2/3 = 6.5833 or 79/12
So the variance is 79/12 – (25/12)2 = 2.2431 or 323/144
Now since standard deviation (σ) is the square root of variance (σ2), it is
√323/144 = 1.4977
Problems 254-258
You are asked for a z-score, so you will use the equation: z = (x- µ)/σ, in which
you are given the mean (µ) of 62 but a variance (σ2) of 25…and we need the
standard deviation; which is fine since we will take the square root of the
variance to get the standard deviation (σ) of 5…and we are ready to plug in the
values for x…
Problems 254
For x = 68, then z = (68 - 62)/5 = 1.20
Problems 255
For x = 49, then z = (49 - 62)/5 = -2.60
Problems 256
For x = 74, then z = (74 - 62)/5 = 2.40
Problems 257
For x = 57.5, then z = (57.5 - 62)/5 = -0.90
Problems 258
For x = 60.2, then z = (60.2 - 62)/5 = -0.36
Problems 259-263
You are asked to take a z-score and give the “…raw score…”, so you will use
take the z-score equation: z = (x- µ)/σ and rearrange it so that it reads: x = z*σ
+ µ; for this equation you are given the mean (µ) of 62 but a variance (σ2) of
25…and we need the standard deviation; which is fine since we will take the
square root of the variance to get the standard deviation (σ) of 5…and we are
ready to plug in the values for z…
Problems 259
For z = 1.7, then x = 1.7*5 + 62 = 70.5
Problems 260
For z = -2.45, then x = -2.45*5 + 62 = 49.75
Problems 261
For z = 3.18, then x = 3.18*5 + 62 = 77.9
Problems 262
For z = 5, then x = 5*5 + 62 = 87
Problems 263
For z = -1.5, then x = -1.5*5 + 62 = 54.5
Problems 264
Words like “Knowing nothing more…” and “…at least what percentage…”,
should tell you the generalness of the answer requested…so this is a
Chebychev problem; and his equation is 1 – 1/k2.
We are given the mean (29) and the standard deviation (4) and a pair of
numbers that the probability must be between (and they MUST be equal
distance from the mean…which they are). Choosing one we calculate the zscore = (37-29)/4 = 2 Now taking that and entering into the Chebychev
equation you get 1 – 1/22 = ¾ or 0.75
So that means that at least 75% of the probability can be found between 21
and 37
Problems 265
“…normally distributed…”, says start with a z-score (unless you want to enter
it directly into your calculator (that will be mentioned at the end)). For 21, zscore = (21-29)/4 = -2, and for 37,
z-score = (37-29)/4 = 2; which means there is a probability of 0.4772 for each
(from the table), giving a total probability of 0.9544
For the TI-83/84, you could go to 2nd-DISTR and down to “normalcdf” and hit
enter. Then enter the following for this problem “21,37,29,4” and hit enter;
[please note that the comma is above the 7 key, and you could close the
parenthesis, but it is not necessary]. This gives an answer of 0.9545 and yes it
is slightly different, but that is because the calculator carries more decimal
places and is not rounded off, so is a little more accurate.
Problems 266-268
“…normally distributed…”, has something to do with z-scores and the
probability table (unless you want to enter it directly into your calculator (that
will be mentioned too)).
Problems 266
For a problem talking about where something “…cuts off…”, you must first find
the amount of probability between the mean (120) and the cutoff
point…which in this case, pertains to the statement “…top 15%...”; and a
picture is always helpful (and recommended) and looks like:
16% or 0.16
120
X
Since the mean cuts the normal probability distribution in half, then there
must be (0.50 – 0.16) 0.3400 probability between the mean and the line
defined as X. I used 4-digits so that it would remind me that in the next step, I
must look in the body of the table and get a z-score (backwards to finding
probabilities)…in this case the closest probability is 0.3389, giving a z-score of
0.99. Now to take to take the z-score and turn it into a raw score, I can either
take the z-score equation (z=(x- µ)/σ ) and plug in all the values except x and
solve for x, or rearrange the equation first (x=zσ +µ) and plug in the numbers:
in either case, x = 0.99(10) + 120 = 129.9
To use the TI-83/84, you press 2nd DISTR and go to “invNorm” and press enter.
Then you enter
“0.84,120,10”; the 0.84 is because that’s how much probability that there is in
the picture from the farthest left point to the line marked X. (Remember the
comma is above the 7 key.)
This gives: 129.9446
Problems 267
For a problem talking about “…probability…”, you have two ways to solve this:
(1) by the use of your calculator (which I will explain afterwards) and (2) by
process, for which you must first find the
z-score(s) at the edges of the “shaded area” (the area you are looking for)
using the equation z=(x- µ)/σ. A picture is always helpful (and recommended)
105
125
and looks like:
The z-score for 105 is z=(105- 120)/10 = -1.50; and for 125 is z=(125- 120)/10 =
0.50. The next step is to use the table and look up the probability values…for
z= -1.50 it is 0.4332 and for z=0.50 it is 0.1915. Since (as the picture shows)
the probability is on both sides of the mean we add the probabilities:
0.4332 + 0.1915 = 0.6247
To use the TI-83/84, you press 2nd DISTR and go to “normalcdf” and press
enter. Then you enter
“105,125,120,10”, (remember the comma is above the 7 key) and hit enter.
This gives: 0.6247