successes, it is a binomial random variable with
10
n
=
and
0.20.
p
=
2.
(
)
(
) (
)
3
7
10
3
0.2
0.8
0.2013.
3
P X
=
=
=
There is a 20% chance that Patti will answer exactly 3
questions correctly.
3.
(
)
(
)
(
)
6
1
6
1
5 .
P X
P X
P X
≥
=−
<
=−
≤
Using binomcdf(10,0.2,5) we get 0.9936.
So
(
)
6
1
0.9936
0.0064.
P X
≥
=−
=
There is only a 0.64% chance that a student would pass so we would be
quite surprised if Patti was able to pass.
Check Your Understanding, page 393:
1.
(
)
10 0.20
2.
X
np
µ
=
=
We would expect the average student to get 2 answers correct on such a
quiz.
2.
(
)
(
)(
)
1
10 0.20
0.80
1.265.
X
np
p
σ
=
−
=
=
We would expect individual students’ scores to vary
from a mean of 2 correct answers by an average of 1.265 correct answers.
3.
Two standard deviations above the mean is
(
)
2
2 1.265
4.53.
+
=
Since Patti can only score an integer
number of questions correctly, we are asked to find
(
)
(
)
5
1
4 .
P X
P X
≥
= −
≤
Using binomcdf(10,0.2,4)
we get 0.9672.
So
(
)
5
1
0.9672
0.0328.
P X
≥
= −
=
There is about a 3% chance that Patti will score
more than two standard deviations above the mean.
Check Your Understanding, page 401:
1. Check the BITS:
Binar
y?
“Success” = roll
doubles.
“Failure” = don’t roll doubles.
Independent:
Rolling dice is an independent process.
Trials:
We are counting trials up to and including the first
success.
Success?
The probability of success is
1
6
for each roll (there are 6 ways to get doubles out of a
total of 36 possible rolls).
This is a geometric setting.
Since
X
is measuring the number of trials to get
the first success, it is a geometric random variable with
1
.
6
p
=
2.
(
)
2
5
1
3
0.1157.
6
6
P T
=
=
=
There is about a 12% chance that you will get the first set of doubles
on the third roll of the dice.
3.
(
)
(
)
(
)
(
)
1
5
1
3
1
2
3
0.1157
0.1667
0.1389
0.1157
0.4213.
6
6
6
P T
P T
P T
P T
≤
=
=+
=+
==+
+
=
+
+
=
Exercises, page 403:
6.69
Binary?
“Success” = seed germinates.
“Failure” = seed does not germinate.
Independent?
Possibly although it is possible that if one seed does not germinate, it’s more likely that others in the
packet will not grow either.
Number?
20 seeds.
Success?
The probability that each seed germinates is
85%, assuming the advertised percentage is true.
Assuming independence does hold, this is a binomial
setting and
X
would have a binomial distribution.
6.70
Binary?
“Success” = name has more than 6 letters.
“Failure” = name has 6 lette
rs or less.
Independent?
Since we are selecting without replacement from a small number of students, the
observations are not independent.
Number?
4 names are drawn.
Success?
The probability that a
148
The Practice of Statistics for AP*
, 4/e
student’s name has more than 6 letters does not change from one draw to the next.
This is a binomial
setting and
Y
would have a binomial distribution.
6.71
Binary?
“Success” = person is left

handed.
“Failure” = person is right
handed.
Independent?
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 Winter '15
 McHugh
 Probability