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Solutions to STAT 515 Homework: Probability Theory, Assignments of Data Analysis & Statistical Methods

University of South Carolina (USC) - ColumbiaData Analysis & Statistical Methods

The solutions to the practice homework for a university-level statistics course, specifically stat 515, focusing on probability theory. It includes calculations for various probability distributions, such as mean and standard deviation, and applications to binomial experiments, including the probability of correct decisions for a psychic with and without esp.

Typology: Assignments

Pre 2010

Uploaded on 09/02/2009

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STAT 515 - Fall 2003 - Solutions to the Practice Homework
Pg. 170: 3.124 a
a) 20! / 10! 6! 4! = 20*19*18*17*16*15*14*13*12*11*10*9*8*7*6*5*4*3*2*1
10*9*8*7*6*5*4*3*2*1 * 6*5*4*3*2*1 * 4*3*2*1
= 20*19*18*17*16*15*14*13*12*11 = 4*19*3*17*4*5*7*13*1*11
6*5*4*3*2*1 * 4*3*2*1 2
= 77597520/2 = 38,798,760
Pg. 186: 4.22 part a only
Using the standard formulas (page 183 and 184) gives
µ
µµ
µ=Σ
ΣΣ
Σx p(x)= 10(0.05)+20(0.20)+30(0.30)+40(0.25)+50(0.10)+60(0.10)
= 0.5+4.0+9.0+10.0+5.0+6.0=34.5
σ
σσ
σ2=Σ
ΣΣ
Σ (x-µ
µµ
µ)2 p(x)
=(10-34.5) 20.05 + (20-34.5) 20.20 + (30-34.5) 20.30 + (40-34.5) 20.25 + (50-34.5) 20.10 + (60-34.5) 20.10
=(-24.5) 20.05 + (-14.5) 20.20 + (-4.5) 20.30 + (5.5) 20.25 + (15.5) 20.10 + (25.5) 20.10
=174.75
σ
σσ
σ=√
σ
σσ
σ2 = √174.75 = 13.21930
Pg. 201: 4.52, also a new part f and g
a) If the psychic is guessing (no ESP) what is the value of p, the probability of a correct decision on each
trial? Only 1 out of 10 boxes is correct, so 1/10.
b) If the psychic is guessing, what is the expected number of correct decisions in seven trials? This is a
binomial experiment with p=0.1 and n=7, so µ
µµ
µ=np=7(0.1)=0.7 correct decisions
c) IF the psychic is guessing, what is the probability of no correct decisions in seven trials? This is a
binomial experiment with p=0.1 and n=7, the problem is asking P(X=0)=(n choose x)px(1-p)n-x
= (7 choose 0)0.10(1-0.1)7 = 1 (1) (0.9) 7 = 0.4782969 You could also answer this one by using just the
multiplication rule, and you would end up with the same (0.9) 7.
d) Now suppose the psychic has ESP with p=0.5. What is the probability of no correct decisions in seven
trials? Same as in c, but with a new p. P(X=0)= (7 choose 0)0.50(1-0.5)7 = 1 (1) (0.5) 7 = 0.0078125
e) If the psychic failed on all seven trials, is this evidence against them having ESP? Explain. Yes. The
chance of them getting none correct if they really have ESP of any worth is only 7 out of a 1,000. That
isn’t very likely!
f) What is the probability that the psychic would get exactly two correct if they had no ESP? This one
would be really annoying to do out the long way, but is easy if you use the binomial formula.
P(X=2)= (7 choose 2)0.12(1-0.1)5 = 7!/(2!5!) (0.1) 2(0.9) 5 = 21 (0.1) 2(0.9) 5 = 0.1240029
g) What is the probability that the psychic would get exactly two correct if they had ESP with p=0.5?
P(X=2)= (7 choose 2)0.52(1-0.5)5 = 21 (0.5) 2(0.5) 5 = 0.1640625

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STAT 515 - Fall 2003 - Solutions to the Practice Homework

Pg. 170: 3.124 a a) 20! / 10! 6! 4! = 201918171615141312111098765432* 10987654321 * 654321 * 432*

= 20191817161514131211 = 419317457131 654321 * 4321 2

= 77597520/2 = 38,798,

Pg. 186: 4.22 part a only Using the standard formulas (page 183 and 184) gives μμμμ=ΣΣΣΣx p(x)= 10(0.05)+20(0.20)+30(0.30)+40(0.25)+50(0.10)+60(0.10) = 0.5+4.0+9.0+10.0+5.0+6.0=34. σσσσ^2 =ΣΣΣΣ (x-μμμμ)^2 p(x) =(10-34.5) 2 0.05 + (20-34.5) 2 0.20 + (30-34.5) 2 0.30 + (40-34.5) 2 0.25 + (50-34.5) 2 0.10 + (60-34.5) 2 0. =(-24.5) 2 0.05 + (-14.5) 2 0.20 + (-4.5) 2 0.30 + (5.5) 2 0.25 + (15.5) 2 0.10 + (25.5) 2 0. =174.

σσσσ= √ σσσσ^2 = √ 174.75 = 13.

Pg. 201: 4.52, also a new part f and g a) If the psychic is guessing (no ESP) what is the value of p, the probability of a correct decision on each trial? Only 1 out of 10 boxes is correct, so 1/10.

b) If the psychic is guessing, what is the expected number of correct decisions in seven trials? This is a binomial experiment with p=0.1 and n=7, so μμμμ=np=7(0.1)=0.7 correct decisions

c) IF the psychic is guessing, what is the probability of no correct decisions in seven trials? This is a binomial experiment with p=0.1 and n=7, the problem is asking P(X=0)=(n choose x)p x(1-p)n-x = (7 choose 0)0.1 0 (1-0.1) 7 = 1 (1) (0.9) 7 = 0.4782969 You could also answer this one by using just the multiplication rule, and you would end up with the same (0.9) 7.

d) Now suppose the psychic has ESP with p=0.5. What is the probability of no correct decisions in seven trials? Same as in c, but with a new p. P(X=0)= (7 choose 0)0.5 0 (1-0.5)^7 = 1 (1) (0.5) 7 = 0.

e) If the psychic failed on all seven trials, is this evidence against them having ESP? Explain. Yes. The chance of them getting none correct if they really have ESP of any worth is only 7 out of a 1,000. That isn’t very likely!

f) What is the probability that the psychic would get exactly two correct if they had no ESP? This one would be really annoying to do out the long way, but is easy if you use the binomial formula. P(X=2)= (7 choose 2)0.1 2 (1-0.1)^5 = 7!/(2!5!) (0.1) 2 (0.9) 5 = 21 (0.1) 2 (0.9) 5 = 0.

g) What is the probability that the psychic would get exactly two correct if they had ESP with p=0.5? P(X=2)= (7 choose 2)0.5 2 (1-0.5)^5 = 21 (0.5) 2 (0.5) 5 = 0.