Homework 9, Math10170, Spring 2015
Question 1 If I roll a six sided die twice and observe the
space is given by:
{(1, 1) (1, 2) (1, 3) (1, 4)
(2, 1) (2, 2) (2, 3) (2, 4)
(3, 1) (3, 2) (3, 3) (3, 4)
(4, 1) (4, 2) (4, 3) (4, 4)
(5, 1) (5, 2) (5, 3) (5, 4)
(6, 1) (6, 2) (6, 3) (6, 4)
numbers on the uppermost faces, the sample
(1, 5) (1, 6)
(2, 5) (2, 6)
(3, 5) (3, 6)
(4, 5) (4, 6)
(5, 5) (5, 6)
(6, 5) (6, 6)}
The random variable X gives the maximum of the pair of numbers. (If the outcome is (2, 3), then X
takes the value 3 and if the outcome is (5, 5), then X takes the value 5 etc... ).
The possible values of X are {1, 2, 3, 4, 5, 6}. (a) What are the possible values of X?
(b) Find the probability distribution for X:
X
1
2
3
4
5
6
P(X) XP(X) (X − µ)
1/36
1/36
-3.47
3/36
6/36
-2.47
5/36 15/36
-1.47
7/36 28/36
-0.47
9/36 45/36
0.53
11/36 66/36
1.53
Total 161/36
(X − µ)2
12.04
6.1
2.16
0.221
0.281
2.341
(X − µ)2 P (X)
0.334
0.508
0.3
0.043
0.073
0.71
1.97
(c) What is the expected value of X (i.e. what is E(X))?
161
≈ 4.47.
36
(d) What is the variance and standard deviation of X (i.e. what is σ 2 (X) and σ(X)?)
E(X) =
Using the approximation µ = E(X) = 4.47, we calculate σ 2 (X) as in the table above.
σ 2 (X) ≈ 1.97
p
σ(X) ≈ σ 2 (X) ≈ 1.4.
1
Question 2 A basketball player with a 70% chance of making a basket takes 4 (independent) shots in
a row. The sample space for this experiment is
{BBBB, BBBM, BBM B, BBM M, BM BB, BM BM, BM M B, BM M M,
M BBB, M BBM, M BM B, M BM M, M M BB, M M BM, M M M B, M M M M }.
Let X denote the number of baskets that the player gets and let Z denote the number of runs (of both
baskets and misses as in the notes) in the outcome.
Outcome
Probability
BBBB
BBBM
BBMB
BBMM
BMBB
BMBM
BMMB
BMMM
MBBB
MBBM
MBMB
MBMM
MMBB
MMBM
MMMB
MMMM
(0.7)4 = 0.2401
(0.7)3 (0.3) = 0.1029
(0.7)3 (0.3) = 0.1029
(0.7)2 (0.3)2 = 0.0441
(0.7)3 (0.3) = 0.1029
(0.7)2 (0.3)2 = 0.0441
(0.7)2 (0.3)2 = 0.0441
(0.7)(0.3)3 = 0.0189
(0.7)3 (0.3) = 0.1029
(0.7)2 (0.3)2 = 0.0441
(0.7)2 (0.3)2 = 0.0441
(0.7)(0.3)3 = 0.0189
(0.7)2 (0.3)2 = 0.0441
(0.7)(0.3)3 = 0.0189
(0.7)(0.3)3 = 0.0189
(0.3)4 = 0.0081
X
=# B’s
4
3
3
2
3
2
2
1
3
2
2
1
2
1
1
0
Z
= # Runs
1
2
3
2
3
4
3
2
2
3
4
3
2
3
2
1
(a) What are the possible values of X and Z?
The possible values of X are {0, 1, 2, 3, 4}.
The possible values of Z are {1, 2, 3, 4}.
(b) Find the probability distributions for X and Z (remember you can use independence to calculate:
P (BM BM ) = (.7)2 (.3)2 etc...
X
0
1
2
3
4
P(X)
0.0081
0.0756
0.2646
0.4116
0.2401
Total
Z
1
2
3
4
P(Z)
0.2482
0.3318
0.3318
0.0882
Total
XP(X) (X − E(X)) (X − E(X))2
0
-2.8
7.84
0.0756
-1.8
3.24
0.5292
-0.8
0.64
1.2348
0.2
0.04
0.9604
1.2
1.44
2.8
ZP(Z) (Z − E(Z)) (Z − E(Z))2
0.2482
-1.26
1.5876
0.6636
-0.26
0.0676
0.9954
0.74
0.5476
0.3528
1.74
3.0276
2.26
2
(X − E(X))2 P (X)
0.0635
0.2449
0.1693
0.0165
0.3457
0.8399
(Z − E(Z))2 P (Z)
0.394
0.0224
0.1817
0.267
0.8581
(c) Find E(X) and E(Z).
E(X) ≈ 2.8 and E(Z) ≈.
(d) FInd the standard deviation of X (σ(X)) and the standard deviation of Z (σ(Z)).
√
√
σ(X) ≈ 0.8399 ≈ 0.92 and σ(Z) ≈ 0.8581 ≈ 0.93.
0
.01
.02
.03
.04
.05
Question 4 The following graph shows two density curves created by Stata from data for 200 to 250
games for the legendary basketball players Michael Jordan and Scottie Pippen (Data collected by Peter
Ulrickson). The variables shown are Z = Points scored per game for M. Jordan and W = Points
scored per game for S. Pippen.
0
10
20
30
40
50
Z, W
Jordan Points Per Game = Z
Pippen Points Per Game = W
(a) Estimate E(Z) and give an interpretation of its meaning.
Recall that the expected value is the balance point of the blue distribution. E(Z) ≈ 29.59 ( your guess
is probably close but not exactly equal to the E(Z)). We have E(Z) equals the average number of
points per game for M. Jordan for the data used.
(b) Estimate E(W ) and give an interpretation of its meaning.
The expected value is the balance point of the red distribution. E(W ) ≈ 19.67. We have E(W ) equals
the average number of points per game for S. Pippen for the data used.
3
0
1
2
3
4
Question 5 The following graph shows three density curves created by Stata from data for 200 to 250
games for the legendary basketball players Michael Jordan and Scottie Pippen and current basketball
player Jeremy Lin (Data collected by Peter Ulrickson). This graph shows the field goal percentage per
game for each player.
0
.2
.4
.6
.8
1
x
fg% per game Lin
fg% per game Pippen
fg% per game Jordan
(a) Estimate the average field goal percentage per game for each player. Which player has the highest
average field goal percentage per game in your estimation?
• For X = FG% per game for M. Jordan, E(X) ≈ 0.48.
• Y = FG% per game for S. Pippen, E(Y ) ≈ 0.46.
• Z = FG% per game for J. Lin, E(Z) ≈ 0.44.
(b) Which player has the greatest standard deviation for field goal percentage per game? (give
reasons for your answer). Which player is the least consistent as far as Field Goal percentage per game
is concerned?
We see that the distribution of field goal percentage per game is most spread out for Jeremy Lin,
therefore, it is the distribution with the greatest standard deviation.
(c) If for each player, we choose a game at random from the data set, estimate the following probabilities
using the area under the density functions above :
1. The probability that M. Jordan’s field goal percentage for the game will be less than 0.3.
This is the area under the red curve to the left of 0.3 which is approximately 0.035 or 3.5%.
2. The probability that S. Pippen’s field goal percentage for the game will be less than 0.3.
This is the area under the green curve to the left of 0.3 which is approximately 0.1 or 10%.
3. The probability that J. Lin’s field goal percentage for the game will be less than 0.3.
This is the area under the blue curve to the left of 0.3 which is approximately .2 or 20%.
4
1
2
3
4
5
Question 4 The following graph shows two density curves created by Stata from data for a football
season for two Quarterbacks; Peyton Manning and Colin Kaepernick. (Data collected by Peter Ulrickson). The variables shown are Completion percentage per game for P. Manning and Completion
percentage per game for C. Kaepernick.
.4
.5
.6
.7
.8
x
Manning
Colin Kaepernick
(a) Estimate the average completion percentage per game for each player. Which player has the highest
average completion percentage per game for the season?
The average completion percentage per game for Peyton Manning from the data is approximately 0.67
or 67%.
The average completion percentage per game for Colin Kaepernick from the data is approximately 0.6
or 60%.
(b) Using the area beneath the above density functions, estimate the following probabilities :
1. The probability that P. Manning’s Completion percentage will be greater than 0.7 for any given
game.
This is the area under the blue curve to the right of 0.7 which is approximately 0.35 or 35%.
2. The probability that C. Kaepernick’s Completion percentage will be greater than 0.7 for any given
game.
This is the area under the red curve to the right of 0.7 which is approximately 0.1 or 10%.
5