Opportunity to Excel # 3
Name:________________________
Find the indicated quantities in problems 1-5. [4 pts. each] - show R code and report numerical answers
1)
2)
3)
P( −1.5 < Z < 2) = 0.9104427
> pnorm(2)-pnorm(-1.5)
[1] 0.9104427
4)
t0.92;5 = 1.649300
> qt(.92,5)
[1] 1.649300
5)
P ( c < Z < 1.4) = 0.65 ⇒ c = −0.615103
> qnorm(pnorm(1.4)-.65)
[1] -0.615103
(
)
P χ 52 > 7 = 0.2206403
> 1-pchisq(7,5)
[1] 0.2206403
P ( − c < t9 < c ) = 0.90 ⇒ c = 1.833113
> qt(.95,9)
[1] 1.833113
Long answers — be sure to1 define your random variable,2 specify the distribution of the random variable, and 3
write a mathematical statement for what the English prose is asking prior to reporting a numerical answer. Write out
all R code used to compute numerical answers.
6)
Find the probability of obtaining 6 or more heads in 10 tosses of a weighted coin, where the probability of
obtaining a head in any given trial is 1/3. [12 pts]
Let X= number of heads in 10 tosses of a weighted coin. [4 pts] X ~ Bin(n = 10, π = 1/ 3) [4 pts]
Want to find P ( X ≥ 6) [2 pts]. Note that P ( X ≥ 6) = 1 − P ( X ≤ 5) = 0.07656353 [ 2pts].
> sum(dbinom(6:10,10,1/3)) # P(X>=6)
[1] 0.07656353
> 1-pbinom(5,10,1/3)
# 1-P(X<=5)
[1] 0.07656353
7)
R- code [2 pts]
Suppose that the population of adult, male black bears has weights that follow as approximate normal
distribution with mean of 350 and standard deviation of 75. What is
a)
the probability that a randomly selected bear weighs more than 500 pounds? [12 pts]
Let X = weight of randomly selected bear. [ 4 pts] X~N(350, 75) [4 pts] Want to find P ( X > 500) [2
pts] P ( X > 500) = 1 − P ( X ≤ 500) = 0.02275013 . [2 pts]
> 1-pnorm(500,350,75)
[1] 0.02275013
b)
#
R- code [2 pts]
the probability the average weight of twenty five randomly selected bears exceeds 380 pounds? [8 pts]
(
)
(
)
(
)
X ~ N 350,75 / 25 [4 pts] - Want to find P X ≥ 380 = 1 − P X ≤ 380 = 0.02275013 .[2 pts]
> 1-pnorm(380,350,75/sqrt(25))
[1] 0.02275013
c)
the 0.75 quantile for bear weights? [ 2 pts]
P ( X ≤ c ) = 0.75 ⇒ c = 400.5867 [2 pts]
> qnorm(.75,350,75)
[1] 400.5867
# R-code [2 pts]
8)
The grades in an evil computer science class follow a continuous uniform [0, 100] distribution. Let the random
variable X be your grade in the unmentionable class. What is:
a) the expected value of X? [3 pts]
a + b 0 + 100
E(X ) =
=
= 50
2
2
b) the variance of X? [5 pts]
σ =
2
X
( b − a ) 2 = (100 − 0) 2
12
12
= 833.3333
c) the standard deviation of X? [5 pts]
σ X = 833.3333 = 28.86751
d) the approximate probability that a class of size n = 49 has a mean less than 54.123930? [9 pts]
σ
28.86751
⎛
⎞
According to the CLT X ~ approxN ⎜ µ X = µ = 50, σ X =
=
= 4.123930⎟ [4 pts] since an n
⎝
⎠
49
n
(
)
of 49 for a uniform is considered sufficiently large. Consequently, P X ≤ 54.12393 = 0.8413447. [3 pts]
> SD <- (100^2/12)^.5
> sigxbar <- SD/sqrt(49)
> pnorm(54.123930,50,sigxbar)
[1] 0.8413447
9)
# R-code [2 pts]
Compute the exact probability of getting 22 or fewer heads in 50 fair coin tosses as well as the approximate
probability given by the normal distribution. Explain why the normal distribution can be used to approximate
the binomial distribution in this problem. [22 pts]
Let X = number of heads in 50 tosses of a fair coin. [4 pts] X ~ Bin(n = 50, π = 1/ 2) [4 pts] Need to find
P ( X ≤ 22) = 0.2399438 . [2 pts]
> pbinom(22,50,1/2)
# R-code [2 pts]
[1] 0.2399438
Since nπ = 50 × 0.5 = 25 > 10 and n (1 − π ) = 50 × 0.5 = 25 > 10 [4 pts], the sampling distribution of X can be
approximated with a normal distribution with mean µ X = n × π = 50 × 0.5 = 25 and standard deviation
σ X = nπ (1 − π ) = 50 × 0.5 × 0.5 = 3.535534 . That is, X ~ approxN (25,3.535534) . Consequently,
⎛ X − 25 22 + .5 − 25 ⎞
≤
P ( X ≤ 22) = P ⎜
⎟ = P ( Z ≤ -0.7071068) = 0.2397501 [4 pts]
⎝ 3.53
3.53 ⎠
> mp <- 50*1/2
> sigp <- sqrt(50*1/2*1/2)
> pnorm(22+.5,mp,sigp)
[1] 0.2397501
# R-code [2 pts]
EXTRA CREDIT 1:
When working with finite populations, the binomial model often becomes untenable. Specifically, when sampling
without replacement, the assumption of constant probability from trial to trial is no longer satisfied. However, deriving
the exact distribution for a finite sample of dichotomous objects is not difficult. Given a dichotomous population of
objects such that m are good and n are bad, the probability of selecting exactly x good items and k − x bad items from a
sample of size k is
⎛ m⎞ ⎛ n ⎞
⎜⎝ x ⎟⎠ ⎜⎝ k − x⎟⎠
.
P ( X = x) =
⎛ m + n⎞
⎜⎝ k ⎟⎠
A computer manufacturer decides to purchase monitors from a new start up company claiming strict quality control
standards. The manufacturer orders 150 monitors and decides to accept the lot provided a random sample of size 25
reveals no defective monitors. If the lot of 150 monitors contains three defective monitors, find
a)
The probability the lot will be accepted?
If X = the number of good monitors, the lot will be accepted when X = 25. Consequently, we need to
find P(X = 25).
⎛147⎞ ⎛ 3 ⎞
⎜⎝ 25 ⎟⎠ ⎜⎝ 25 − 25⎟⎠
P ( X = 25) =
= 0.576365
⎛147 + 3⎞
⎜⎝ 25 ⎟⎠
> choose(147,25)*choose(3,0)/choose(150,25)
[1] 0.576365
> # OR Using R dhyper()
> dhyper(25,147,3,25)
[1] 0.576365
b)
The expected value of X.
E ( X ) = ∑ x × P ( X = x ) = 24.5
c)
> x <- 22:25
> m <- 147
> n <- 3
> k <- 25
> px <- choose(m,x)*choose(n,(k-x))/choose((m+n),k)
> px
[1] 0.004171957 0.068021041 0.351442046 0.576364956
> dhyper(x,m,n,k)
[1] 0.004171957 0.068021041 0.351442046 0.576364956
> EX <- sum(x*px)
> EX
[1] 24.5
The variance of X.
σ X2 = E ( X − µ X ) = ∑ ( x − µ X ) × P ( X = x ) =0.4110738
2
2
> VX <- sum(((x-EX)^2)*px)
> VX
[1] 0.4110738
EXTRA CREDIT 2:
Problem 6.7 from your homework stated that when you add any number of independent squared standard normals, the
resulting distribution is a chi-square distribution with degrees of freedom equal to the number of independent squared
standard normals you add together. There is a simple relationship between the mean, variance, and the degrees of
freedom for a chi-square distribution. What is this relationship? You might use simulation to investigate the
relationships. The mean and variance of a chi-square distribution with ν degrees of freedom are ν and 2ν
respectively.