Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
Homework 6 Solutions
60a. We are told to use the exponential distribution with = .01386
100
page 1
Interpret probability that the distance is at most 100 as :
( 100) =
0
Similarly, at most 200 is:
( 200) =
200
0
() = (100) = 1 100 = 1 1.386 = .
() =
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
Homework 7 Solutions
page 1
22a. We are given the probability table below. Now we want to know the expectation for the recorded score (x
( + )(, ) = (0 + 0) .02 + (0 + 5) .06 + (0 + 10) .02 + (0 + 15) .10 + (5 + 0)
+ y), which is given by:
.04 + (5 + 5)
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
Homework 10 Solutions
page 1
18 (321)
a. Under the null hypothesis, = 75 = 9, so the test statistic is =
(72.375)
=
2.7
9/5
= 1.5 .
The observed mean of 72.3 is 1.5 standard deviations below the null value (border value of the mean
under the null hypothes
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
Homework 11 Solutions
page 1
37 (327) Since we have large sample sizes and our proportions are away from 0 and 1, we can use the normal
approximation to construct z statistics. Proceed as follows:
BMI Group
(0,25)
[25,30)
Count
262
159
sample proportion
0
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
1. (a) The arrival rate for 5book orders is 5(0.1) = 0.5 per minute. So the
probability of no arrivals of this type in 1 minute is e0.5 0.6065.
(b) The interorder time T exponential(5), so the required probability is
10/60
5e5x dx, which is 1 e5/6 , or
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
1. (a)
4
0
4
1
1
2
4
2
4
3
3
4
3
(b)
G=1+
So
4 4 4 4 4 4 4 4 4 4
1, , , ,
1 1 2 1 2 3 1 2 3 3
9 36 72 96 128
=
,
,
,
,
.
341 341 341 341 341
This is the probability that the next transition is to state 4. The probability is 4/(3 + 4), or 4/7.
4
This is t
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
Homework 3 Solutions
page 1
P( ) = P()2 =. 452 = 0.2025
74. The importance of having independent events is that we can multiply the respective probabilities.
Since we want the probability of a match, there are 4 ways of matching (both having type A, type
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
Homework 5 Solutions
page 1
6a.
( 3)3 4

3
2
b. We know that the pdf must integrate to 1 over the range of X.
4
1 = 1 ( 3)2 =
Then k = 3/4
c.
2
4
( > 3) = .75 1 ( 3)2 = .75
3
1 1
= 2 = 4/3
3 3
( 3)3 4

3
3
=
3
1
1 =
4
3
0.5
For these next few parts, i
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
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Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
Homework 9 Solutions
page 1
3. This question tests your understanding of the definition of a confidence interval. We are told that a 95%
confidence interval for the average alcohol content for the population of all bottles is (7.8 , 9.4).
two extremes: a
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
Homework 2 Solutions
page 1
2ad. . Each car can go three possible directions, and there are three cars, so there are 3*3*3 = 27 total choices.
all choices are listed below
= cfw_RRR, RRL, RRS,
RLR, RLL, RLS
RSR, RSL, RSS,
SRR, SRL, SRS,
SLR, SLL, SLS,
SS
Probability & Statistics for the Physical Sciences
ST 380

Spring 2014
Chapter 1 Solutions
page 1
4a. Three examples of concrete populations
i. Basketball athletes currently at NCSU
ii. Sections of a specific concrete floor
iii. Mosquitoes in the Raleigh area
Three examples of hypothetical populations
i. All possible toxicit