Download Quiz Seven Solutions

May 29, 2009
Math 244 Spring 2009 Quiz 7
Name:
1. Suppose sample data is collected for a particular measurement. The resulting 15 measurements are:
0.68, 2.1, 3.6, 5.3, 7.1, 9.1, 11.4, 13.9, 16.7, 20, 24.1, 29.1, 35.8, 46, 68
Construct two probability plots, one comparing the data with a normal distribution and another comparing it to a exponential distribution. You may do this by hand or with a computer
but show your calculations clearly and make your plot as accurate as possible. Then decide
which distribution better describes the data. For the normal probability plot you could construct a table with the percentiles and the corresponding values from the standard normal
distribution. I used a more sophistocated calculation tool to get the values rather than the
table but your values should agree with the normal values to two decimal digits if you used
the table.
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Percentile in Decimal From
0.033333333
0.1
0.166666667
0.233333333
0.3
0.366666667
0.433333333
0.5
0.566666667
0.633333333
0.7
0.766666667
0.833333333
0.9
0.966666667
Stnd. Norm Values
-1.833914636
-1.281551566
-0.967421566
-0.727913291
-0.524400513
-0.340694827
-0.167894005
-1.39146E-16
0.167894005
0.340694827
0.524400513
0.727913291
0.967421566
1.281551566
1.833914636
Observation
0.68
2.1
3.6
5.3
7.1
9.1
11.4
13.9
16.7
20
24.1
29.1
35.8
46
68
The resulting probability plot plots the third column against the fourth column and gives:
Normal
Probability
Plot
Normal
Probability
Plot
80
70
60
50
40
30
20
10
0
‐2.5
‐2
‐1.5
‐1
‐0.5
0
0.5
1
1.5
2
2.5
For the exponential comparison, the only column that changes is the Target. You can utelize
the fact that the cdf for an exponential random variable with λ = 1 is easily written down as
F (x) = 1 − e−λx .
i
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Percentile in Decimal From
0.033333333
0.1
0.166666667
0.233333333
0.3
0.366666667
0.433333333
0.5
0.566666667
0.633333333
0.7
0.766666667
0.833333333
0.9
0.966666667
The resulting probability plot looks like:
Exponential Values
0.033901552
0.105360516
0.182321557
0.265703166
0.356674944
0.456758402
0.567984038
0.693147181
0.836248024
1.003302109
1.203972804
1.455287233
1.791759469
2.302585093
3.401197382
Observation
0.68
2.1
3.6
5.3
7.1
9.1
11.4
13.9
16.7
20
24.1
29.1
35.8
46
68
Exponen'al
Probability
Plot
Exponen2al
Probability
Plot
80
70
60
50
40
30
20
10
0
0
0.5
1
1.5
2
2.5
3
3.5
4
It is clear that the second plot is much more linear then the first plot, thus the data fits closely
to what would be expected with an exponential random variable when sampled 15 times.
2. Answer problem number 41 from chapter 5 (page 212). ”Let X be the number of packages
being mailed by a randomly selected customer at a certain shipping facility. Suppose the
disribution of X is as follows:
x
1 2 3 4
p(x) .4 .3 .2 .1
(a) Consider a random sample of size n=2 (two customers), and let X be the sample mean
number of packages shipped. Obtain the probability density distribution of X
We could start by building a table, here I will also include the values of r used for part
c as well.
x1
1
1
1
1
2
2
2
2
3
3
3
3
4
4
4
4
x2
1
2
3
4
1
2
3
4
1
2
3
4
1
2
3
4
p(x1 , x2 )
.16
.12
.08
.04
.12
.09
.06
.03
.08
.06
.04
.02
.04
.03
.02
.01
x
1
1.5
2
2.5
1.5
2
2.5
3
2
2.5
3
3.5
2.5
3
3.5
4
R
0
1
2
3
1
0
1
2
2
1
0
1
3
2
1
0
From this it is pretty clear that the pmf for the mean X will be:
x
1
1.5
2
2.5
3
3.5
4
pX (x)
.16
.24
.25
.2
.1
.04
.01
(b) Refer to part (a) and calculate P (X ≤ 2.5).
For this probability we just need the to add up the probability for all the cases where
x ≤ 2.5. Namelu:
P (X ≤ 2.5) = .16 + .24 + .25 + .2 = .85
(c) Again consider a random sample of size n = 2, but no focus on the statistic R = the
sample range (difference between the largest and smallest values in the sample). Obtain
the distribution of R.
Using the table from (a) we have that the pmf for R is:
r
0
1
2
3
pR (r)
.3
.4
.22
.08
(d) If a random sample of size n = 4 is selected, what is P (X ≤ 1.5?
To answer this question I am going to make a table for the values of X1 , X2 , X3 , and
X4 , but only include cases where the mean is less than or equal to 1.5.
x1
1
2
1
1
1
2
2
2
1
1
1
3
1
1
1
x2
1
1
2
1
1
2
1
1
2
2
1
1
3
1
1
x3
1
1
1
2
1
1
2
1
2
1
2
1
1
3
1
x4
1
1
1
1
2
1
1
2
1
2
2
1
1
1
3
p(x1 , x2 , x3 , x4 )
(.4)4
(.4)3 (.3)
(.4)3 (.3)
(.4)3 (.3)
(.4)3 (.3)
(.4)2 (.3)2
(.4)2 (.3)2
(.4)2 (.3)2
(.4)2 (.3)2
(.4)2 (.3)2
(.4)2 (.3)2
(.4)3 (.2)
(.4)3 (.2)
(.4)3 (.2)
(.4)3 (.2)
x
1
1.25
1.25
1.25
1.25
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
1.5
Since these are the only cases then the probability P (X ≤ 1.5) = (.4)4 + 4(.4)3 (.3) +
6(.4)2 (.3)2 + 4(.4)3 (.2) = .24
3. Answer problem number 50 from chapter 5 (p 218)
The breaking strength of a rivet has a mean value of 10,000 psi and a standard deviation of
500 psi.
(a) What is the probability that the sample mean breaking strength for a random sample
of 40 rivets is between 9900 and 10,200?
Since we are taking a simple random sample with n = 40 > 30 then the Central Limit
Theorem applies and the sample mean X√is approximated well by a normal random
variable with µX = 10, 000 and σX = 500/ 40 ≈ 79.0569. We use this fact to carry out
the calculation:
P (9900 ≤ X ≤ 10, 200) ≈ P
= Φ
9900 − 10000
500
√
40
≤Z≤
10200 − 10000
√ !
√ !
40
2 40
−Φ −
5
5
!
500
√
40
= Φ (2.53) − Φ (−1.26)
≈ .9943 − .1038 = .8905
(b) If the sample size had been 15 rather than 40, could the probability requested in part
(a) be calculated from the given information.
No, in this case since the distribution of an individual rivet is not given then we can not
compute the behavior of X unless we have n large enough to invoke the Central Limit
Theorem. Unfortunately we can not do that here since n = 15 < 30.