MAT 320 Test # 2
1.
a.
b.
c.
12-November-2008
Name:
A large assembly line produces working items with probability of 0.75.
Suppose a random sample of 6 items is taken from the assembly line.
Find the probability that:
All 6 items are working.
At least 1 of the 6 items is working
At least 2 of the 6 items are working.
X = number of working parts in the sample. X ~ Bin(n = 6, p = .75)
a.
P(X = 6) = .178
b.
P(X ≥ 1) = 1 – P(X = 0) = .9998
c.
P(X ≥ 2) = 1 – P(X ≤ 1) = 1 - .005 = .995
2.
A copy machine is reported to break down on average 0.8 times per
month. Assume the number of copier breakdowns follows a Poisson
distribution.
a. Find the probability that the copier breaks down exactly 2 times in a month.
b. Find the probability the copier breaks down more than 2 times in a month.
c. Find the mean and standard deviation of copier breakdowns in a YEAR.
Y = number of breakdowns in a month
Y~ Poisson(λ = .8)
a.
P(Y = 2) = .144
b.
P(Y > 2) = 1 – P(Y ≤ 2) = 1 - .953 = .047
c.
W ~ Poisson(λ = 9.6) E(W) = 9.6 S(W) = √9.6 = 3.098
3.
Joe rolls a fair die until he rolls a 6.
a.
Find the probability that he rolls the die at least twice?
b.
Find the probability that he rolls the die at least three times?
X = number of non sixes
X ~ NB(p = 1/6, r = 1)
Number of rolls = X + 1
a.
P(X ≥ 1) = 1 – P(X = 0) = 1 – 1/6 = 5/6 = .833
b.
P(X ≥ 2) = 1 – P(X ≤ 1) = 1 – [1/6 + 5/36] = 1 = 11/36 = 25/36 = .694
4.
Let the continuous random variable X have the following pdf.
kx2 0  x  3 
f ( x)  

 0 otherwise 
a.
Find k.
b.
Find the cdf of X.
3
3
a.
kx3
1   kx dx 
 9k  k = 1/9
3 0
0
2
b.
x
x<0
0≤x≤3
x≥3
F(x)
0
x3/27
1
x
x
1 2
t3
x3
F ( x)   t dt 

9
27 0 27
0
5.
The developing time of photographs is a random variable having a
normal distribution with a mean of 13.40 seconds and a standard
deviation of 0.68 seconds.
a. Find the probability the developing time takes longer than 14 seconds.
b. Find the probability the developing time takes between 14 and 15 seconds.
c. Find t, the time such that 5% of the developing times are more than t.
X ~ N(μ = 13.4, σ = .68)
a.
P(X > 14) = P(Z > .88) = .1894
b.
P(14 < X < 15) = P(Z < 2.35) – P(Z < .88 ) = .9906 - .8106 = .1800
c.
t = 1.645(.68) + 13.4 = 14.5186
6.
The lifetime of a special transistor is given by the random variable X (in
weeks). X has a Gamma distribution with mean = 12 and variance = 72.
a.
Find the pdf of X.
b.
Find the probability that the lifetime is less than 24 weeks.
a.
E(X) = αβ = 12
V(X) = αββ = 72
β=6
α=2
f ( x) 
1
xe  x / 6
36
24
b.
P(X < 24) =
1
 36 xe
x / 6
for x > 0 and f(x) = 0 otherwise
dx = .908
0
convert to Incomplete gamma function,
Y ~ Standard Gamma(α = 2) P(X < 24) = P(Y < 4) = .908
7.
Let a and b be real constants and X be a discrete random variable such
that the E(X) = μ. Using the definition of expected value, prove that:
E(aX + b) = aμ + b.
{Hint: E[h(X)] = Σh(x)p(x)}
Proof:
E(aX + b) = Σ(ax + b)p(x) = Σ((ax)p(x) + b(p(x))) =
a Σ [x p(x)] + b Σ p(x) = aμ + b
because Σ [x p(x)] = E(X) = μ and Σ p(x) = 1 (pmf).
8.
Let X have an exponential distribution with parameter λ.
Let t > 0 and let c > 0.
Prove that P(X > t) = P(X > c + t | X > c)
Proof:
P(X > t) = exp(-λt)
P(X > c + t | X > c) = P(X > c + t ∩ X > c) / P(X > c)
but P(X > c + t ∩ X > c) = P(X > c + t)
P(X > c + t | X > c) = P(X > c + t) / P(X > c)
P(X > c + t | X > c) = exp(-λc – λt) / exp(-λc)
P(X > c + t | X > c) = exp(-λt) = P(X > t)
9.
The working lifetime of a certain type of circuit has a Uniform
distribution with minimum of 50 and maximum 150 days. A computer
has 4 of these circuits and needs at least 1 of the circuits to be ‘alive’ in
order to function. Assume that the lifetime of each circuit is independent
and they were all installed at the same time.
a. What is the probability any one circuit lasts longer than 125 days?
b. What is the probability the computer functions longer than 125 days?
X = lifetime of a circuit.
X ~ Uniform(50, 150)
f(x) = .01 when 50 < X < 150
a.
P(X > 125) = 25 * .01 = .25 = ¼
Y = number of the 4 circuits that lasts longer than 125 days
Y ~ Bin(n = 4, p = .25)
b.
P(Y ≥ 1) = 1 – P(Y = 0) = 1 – 0.316 = .684
10.
Bolt thread length is normally distributed with mean μ and standard
deviation σ. A bolt is randomly selected. Find the probability that:
a. the thread length is within 1.5 standard deviations of its mean.
b. the thread length is between 1 and 2 standard deviations of its mean.
Y ~ N(μ , σ)
a.
P(-1.5σ < Y – μ < 1.5σ) = P(Z < 1.5) – P(Z < -1.5) =
.9332 - .0668 = .8664
b.
P(-2σ < Y – μ < -1σ) + P(1σ < Y – μ < 2σ)
= P(-2 < Z < -1) + P(1 < Z < 2) =
= 2 * (.9772 - .8413) = 2 * (.1359) = .2718