Quiz 3
Lucy D’Agostino McGowan
September 9, 2014
Q1 Rosner’s Ch 5 Nutrition set of problems, 5.6 - 5.9.
5.6
140−124
20
=
16
20
= .8
1-pnorm(.8)
## [1] 0.2119
21.2%
5.7
90−124
20
= −1.7
1-pnorm(1.7)
## [1] 0.04457
4.46%
5.8
140−121
19
=1
1-pnorm(1)
## [1] 0.1587
15.9%
5.9
90−121
19
= −1.63158
1-pnorm(1.63158)
## [1] 0.05138
5.2%
Q2 Rosner’s Ch 5 Nutrition set of problems, 5.21 - 5.24.
5.21
2.5−3.5
0.6
=
−1
0.6
= -1.667
1
pnorm(-1.66667)
## [1] 0.04779
#or
pnorm(2.5,3.5,0.6)
## [1] 0.04779
5.22
2.5−4
0.5
=
−1.5
0.5
= -3
pnorm(-3)
## [1] 0.00135
#or
pnorm(2.5,4,0.5)
## [1] 0.00135
5.23
2.5−(4−0.03∗(30))
0.02∗30
=
2.5−3.1
0.6
= -1
pnorm(-1)
## [1] 0.1587
#or
pnorm(2.5,4-0.03*30,0.02*30)
## [1] 0.1587
5.24
5.23
2.5−(4−0.03∗(50))
0.02∗50
=
2.5−2.5
0.6
=0
pnorm(0)
## [1] 0.5
#or
pnorm(2.5,4-0.03*50,0.02*50)
## [1] 0.5
Q3a Let X ~ Binom(20, 0.17). Let Y ~ N( mu=E[X], sigma=sqrt(V[X]) ), i.e. Y is Normal with mean and
variance equal to that of the Binomial X. Take a large sample from each of X and Y and sort them from
smallest to largest. Plot Y by X. By large, I mean try getting samples of 10ˆ6. If that crashes your laptop,
go with 10ˆ5.
2
−5
0
y
5
10
x<-rbinom(10^6,20,.17)
mean.x <- 20*.17
sd.x <- sqrt(20*.17*(1-.17))
y<-rnorm(10^6,mean.x, sd.x)
x<-sort(x)
y<-sort(y)
plot(x,y)
0
2
4
6
8
x
Q3b Using the samples from 3, estimate the following probabilities:
q3b1 <- mean.x+sd.x
q3b2 <- mean.x+(2*sd.x)
q3b2.5 <- mean.x+(2.5*sd.x)
q3b3 <- mean.x+(3*sd.x)
• P( X > E[X] + 1*sqrt(V[X]) )
sum(x > q3b1)/10^6
## [1] 0.1095
• P( Y > E[X] + 1*sqrt(V[X]) )
3
10
12
sum(y > q3b1)/10^6
## [1] 0.1587
• P( X > E[X] + 2*sqrt(V[X]) )
sum(x > q3b2)/10^6
## [1] 0.04073
• P( Y > E[X] + 2*sqrt(V[X]) )
sum(y > q3b2)/10^6
## [1] 0.02286
• P( X > E[X] + 2.5*sqrt(V[X]) )
sum(x > q3b2.5)/10^6
## [1] 0.01251
• P( Y > E[X] + 2.5*sqrt(V[X]) )
sum(y > q3b2.5)/10^6
## [1] 0.006184
• P( X > E[X] + 3*sqrt(V[X]) )
sum(x > q3b3)/10^6
## [1] 0.003214
• P( Y > E[X] + 3*sqrt(V[X]) )
sum(y > q3b3)/10^6
## [1] 0.001302
Q4a Repeat Q3a for Binom(100, 0.17).
x<-rbinom(10^6,100,.17)
mean.x <- 100*.17
sd.x <- sqrt(100*.17*(1-.17))
y<-rnorm(10^6,mean.x, sd.x)
x<-sort(x)
y<-sort(y)
plot(x,y)
4
30
20
0
10
y
5
10
15
20
x
Q4b Repeat Q3b for Binom(100, 0.17).
q3b1 <- mean.x+sd.x
q3b2 <- mean.x+(2*sd.x)
q3b2.5 <- mean.x+(2.5*sd.x)
q3b3 <- mean.x+(3*sd.x)
• P( X > E[X] + 1*sqrt(V[X]) )
sum(x > q3b1)/10^6
## [1] 0.1747
• P( Y > E[X] + 1*sqrt(V[X]) )
sum(y > q3b1)/10^6
## [1] 0.1587
• P( X > E[X] + 2*sqrt(V[X]) )
sum(x > q3b2)/10^6
5
25
30
35
## [1] 0.02712
• P( Y > E[X] + 2*sqrt(V[X]) )
sum(y > q3b2)/10^6
## [1] 0.02262
• P( X > E[X] + 2.5*sqrt(V[X]) )
sum(x > q3b2.5)/10^6
## [1] 0.008045
• P( Y > E[X] + 2.5*sqrt(V[X]) )
sum(y > q3b2.5)/10^6
## [1] 0.006088
• P( X > E[X] + 3*sqrt(V[X]) )
sum(x > q3b3)/10^6
## [1] 0.00205
• P( Y > E[X] + 3*sqrt(V[X]) )
sum(y > q3b3)/10^6
## [1] 0.001265
Q5a Repeat Q3a for Binom(1000, 0.17).
x<-rbinom(10^6,1000,.17)
mean.x <- 1000*.17
sd.x <-sqrt(1000*.17*(1-.17))
y<-rnorm(10^6,mean(x), sd(x))
x<-sort(x)
y<-sort(y)
plot(x,y)
6
200
120
160
y
120
140
160
180
x
Q5b Repeat Q3b for Binom(1000, 0.17).
q3b1 <- mean.x+sd.x
q3b2 <- mean.x+(2*sd.x)
q3b2.5 <- mean.x+(2.5*sd.x)
q3b3 <- mean.x+(3*sd.x)
• P( X > E[X] + 1*sqrt(V[X]) )
sum(x > q3b1)/10^6
## [1] 0.1657
• P( Y > E[X] + 1*sqrt(V[X]) )
sum(y > q3b1)/10^6
## [1] 0.1586
• P( X > E[X] + 2*sqrt(V[X]) )
sum(x > q3b2)/10^6
7
200
220
## [1] 0.02538
• P( Y > E[X] + 2*sqrt(V[X]) )
sum(y > q3b2)/10^6
## [1] 0.02263
• P( X > E[X] + 2.5*sqrt(V[X]) )
sum(x > q3b2.5)/10^6
## [1] 0.007375
• P( Y > E[X] + 2.5*sqrt(V[X]) )
sum(y > q3b2.5)/10^6
## [1] 0.006216
• P( X > E[X] + 3*sqrt(V[X]) )
sum(x > q3b3)/10^6
## [1] 0.001711
• P( Y > E[X] + 3*sqrt(V[X]) )
sum(y > q3b3)/10^6
## [1] 0.00131
Q6 Discuss what Q3 - Q5 teaches us. A bullet point discussion is fine.
As we increase the number of trials, we see that the binomial distribution better approximates the normal
distribution
8