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1. The average profit per unit for Product A is $5.32, Product B is $7.98,
and Product C is $9.69. If the percentages of each sold are 36, 29, and 35
respectively, what is the weighted mean profit per unit?
Solution. The weighted mean profit per unit is
5.32*36%+7.98*29%+9.69*35%=1.9152+2.3142+3.3915=7.6209
2. A fleet of cabs for a small cab company consists of five units. The
number of cabs in service varies over time and is a random variable. Daily
log records show the probability distribution for this random variable as
given below:
# in Service
x
0
1
2
3
4
5
Probability
P(x)
.05
.15
.16
.25
.21
.18
Find the mean and standard deviation of number of units in service.
Solution. Denote the mean and standard deviation of X by E(X) and D(X)
respectively. Then, by a formula of E(X)=  kP( X  k ) we have
E(X)=0*0.05+1*0.15+2*0.16+3*0.25+4*0.21+5*0.18=0.15+0.32+0.75+0.8
4+0.9=2.96
By a formula of E(X^2)=  k 2 P( X  k )
=0*0.05+1*0.15+4*0.16+9*0.25+16*0.21+25*0.18
=0.15+0.64+2.25+3.36+4.5=10.9
So, D(X)= E ( X 2 )  [ E ( X )]2
= 10.9  2.962 =1.462326913
3. The Southway National Bank surveyed the status of student accounts
and found that the average overdraft was $21.22 with a standard
deviation of $5.49. If the distribution is normal, find the probability
of a student being overdrawn by more than $18.75.
Solution. Let X be such random variable. By the hypothesis,
X~N(21.22, 5.49^2). So the probability
Pr(X>18.75)=Pr((X-21.22)/5.49>-0.45)=0.674
Note: Y=(X-21.22)/5.49 follows the Standard Normal Distribution(SND),
so by looking up the SND function table, you can get the above probability.
4. An efficiency expert makes periodic checks for weighting errors for a
long distance shipping firm. The expert inspects for errors in
weighing, in recording weights, and errors in processing the bills of
lading. Based on past records the number of weekly errors for all
shipments averages 5.3 with a standard deviation of 1.23 and the
frequency histogram approximates a normal distribution. Suppose x
is the number is the number of weighing errors that will occur next
week. Compute the approximate probability for x=3.
Solution. Let X be the number of weekly errors, By hypothesis, X follows
approximately a normal distribution , ie., X~N(5.3, 1.23^2).
Now we can compute the probability of X>=x, ie., Pr(X>=x). For example,
x=3,
Pr(X>=3)=Pr((X-5.3)/1.23>=-1.87)
=0.96926
Note: Y=(X-5.3)/1.23 follows the Standard Normal Distribution(SND), so
by looking up the SND function table, you can get the above probability.
You can use website:
http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/stat/normaldist.ht
ml
To compute the above probability.
5. An elementary school teacher learned that 40% of school age children
have at least 3 cavities.
a) If the standard deviation is 2.684, how many students would she
expect to find her class of 30 who have at least 3 cavities?
Determine the probability that more than 20 students in her class will have 3
cavities.
Solution. Assume that there are n students in the class. Then let X be the
number of the students who have at least 3 cavities. Obviously, X is a
random variable and follows binomial distribution X~b(n,0.4). We know
that the standard deviation of X is
n * 0.4 * (1  0.4)
So, n * 0.4 * (1  0.4) =2.684
So, n=30.01606667
We choose an integer close to 30.01606667, and we know it should be 30.
Thus, n=30
The expected value of X is E(X)=n*0.4=30*0.4=12.
The formula for the mean and standard deviation of binomial distribution
X~b(n,p) is E(X)=np and D(X)^2=np(1-p)
The probability that more than 20 students in her class will have 3 cavities is
Pr(X>20)=Pr(X=21)+Pr(X=22)+…+Pr(X=30)
 30 
 21 
 30 
 * 0.4 22 * 0.68 +…+ 0.430
 22 
=   * 0.4 21 * 0.69 + 
 30 
k 
Since Pr(X=k)=   0.4 k 0.630  k , k=0,1,…,30