Download Beals ch6 exercises

6.1 If R is a binomial random variale with mean 300 and standard
deviation 15, what are the values of n and p?
np=300 and np(1-p)=225 or 300(1-p)=225 or 300-300p=225
300p=300-225=75 or p=75/300=0.25 and n=300/0.25=1200
6.3. A blindfolded person is given several sips of each of three cups of
cola. He is asked to state for each cup whether it is brand A or
brand B.
a) If in fact his answers are determined by chance alone, what is the
probability that his answers will all be correct?
b) If the same test is given to 100 persons, what is the probability
that at least 10 will give all correct answers? (Use the normal
approximation to the binomial distribution.)
for (a) dbinom(3,3,0.5)=1/8=0.125
for (b) p (z > -0.907)
p=0.125 and n=100
1-pnorm((9.5-12.5)/3.307)=0.8178407
exact:
1-pbinom(9,n,p) = 0.8163046
6.5 A certain type of electric light bulb has a mean life time of 1500
hours and a standard deviation of 150 hours. Four bulbs are connected
so that when one burns out, another will go on. Assuming that the
lifetime of each bulb is independent of the others and is normally
distributed, what is the probability that lighting will continue for
a) at least 6300 hours?
b) at most 5550 hours?
a) 0.1587 [(P(z>1) where z=(6300-6000)/sqrt(4*150^2)]
b) 0.0668 [P(z<-1.5) where z=(5550-6000)/sqrt(4*150^2)
6.7 Suppose the probability distribution of I.Q. scores is normal, with
mean of 100 and variance of 225.
a) What is the probability that a randomly chosen individual has
I.Q. as high as 130?
b) What percent of I.Q. scores are between 91 and 97?
c) What percent of I.Q. scores are between 85 and 103?
e) What percent are less than 106?
a)1-pnorm((130-100)/sqrt(225))=0.0228
b)pnorm((97-100)/sqrt(225))-pnorm((91-100)/sqrt(225))=0.1464 (14.64%)
c)pnorm((103-100)/sqrt(225))-pnorm((85-100)/sqrt(225))=0.4206 (42.06%)
d)pnorm((106-100)/sqrt(225))=0.6554 (65.54%)
6.9. Booklets are packaged in bundles of 100 by weighing them. Suppose
the weight of a booklet is considered to be a random variable with
mean 1 ounce and standard deviation 0.04 ounce. If a bundle weighs
100 ounces to the nearest ounce, it is assumed to contain 100 booklets
and is moved on as is for mailing. If it weighs less or more than
100 ounces, booklets are added or removed until the weight is 100
ounces (to the nearest ounce).
a) What is the probability that a bundle containing 100 booklets is
accepted as is by the weighing procedure?
b) What is the probability that a bundle containing 101 booklets is
accepted as is by the weighing procedure?
a)pnorm((100.5-100)/sqrt(100*0.04^2))-pnorm((99.5-100)/sqrt(100*0.04^2))=0.7888
= P(Z<1.25) – P(Z<-1.25)
b)pnorm((100.5-101)/sqrt(101*0.04^2))-pnorm((99.5-101)/sqrt(101*0.04^2))=0.106692
=P(Z<-1.243796)-P(Z<-3.731389)
if you use var=100*0.04^2 then the answer is 0.1056
pnorm((100.5-101)/sqrt(100*0.04^2))-pnorm((99.5-101)/sqrt(100*0.04^2))=
0.1055614
6.13. Two series of numbers are divided term-by-term and each quotient
is rounded off to the nearest integer value. The round-off error may
be regarded as a continuous random variable uniformly distributed
between -0.5 and 0.5.
a) What is the expected value of the round-off error? What is the
variance of the round-off error?
b) If n of the quotients are averaged, what is the expected value of
the mean round-off error? What is the variance of this sample mean?
a)mean =0, var=1/12 (this is a continuous uniform distribtion with f(x)=1
and between -0.5 and 0.5. the var(x)=E(x^2) since the mean=0.
E(x^2)=integrate x^2=x^3/3 from -0.5 to 0.5 = (x^3/3) from -0.5 to 0.5
= ((1/2)^3)/3+((1/2)^3)/3=1/12
b) mean=0; var(x_bar)=var(x)/n= 1/12n