Let X be the number of packages being mailed by a random customer at a certain s
ID: 3122851 • Letter: L
Question
Let X be the number of packages being mailed by a random customer at a certain shipping facility, and that customers only send between one and four packages at a time. 25% of all customers send one package, 40% send two packages, 25% send three packages, and the rest send four packages. Consider a random sample of two customers and let bar X be the sample mean number of packages shipped a) obtain the probability distribution of bar X. b) Use your answer in part a to calculate P(bar X lessthanorequalto 3). c) What is the expected value of bar X? d) What is the variance of bar X?Explanation / Answer
Ans-
A) They are asking for probability distribution of Xbar for a sample size n=2(two customers). We can calculate the probabilities for Xbar by:
•
n X X i = • In words the Xbar is the average of all values X i.
Table 1 shows the outcomes and probabilities for Xbar of n=2(two customers). The table is set up so we can look at all the possibilities of having two customers (i.e.- if the first customer is shipping one package then the second customer can be shipping one, two, three, or four packages). This could become very monotonous if we had to look at a large number of customers. We can assume independence because we are randomly selecting customers, therefore the probability of Xbar can be calculated by: • () ()() 12 xpxpxp =
Jeff Wetzel Engr 323 Problem 5.41
TABLE: 1 x1 p(x1) x2 p(x2) p(x1,x2) xbar p(xbar) 1 0.4 1 0.4 (1,1) (x1+x2)/2=1.0 .04*.04=.16 1 0.4 2 0.3 (1,2) 1.5 0.12 1 0.4 3 0.2 (1,3) 2.0 0.08 1 0.4 4 0.1 (1,4) 2.5 0.04 2 0.3 1 0.4 (2,1) 1.5 0.12 2 0.3 2 0.3 (2,2) 2.0 0.09 2 0.3 3 0.2 (2,3) 2.5 0.06 2 0.2 4 0.1 (2,4) 3.0 0.03 3 0.2 1 0.4 (3,1) 2.0 0.08 3 0.2 2 0.4 (3,2) 2.5 0.06 3 0.2 3 0.2 (3,3) 3.0 0.04 3 0.2 4 0.1 (3,4) 3.5 0.02 4 0.1 1 0.4 (4,1) 2.5 0.04 4 0.1 2 0.3 (4,2) 3.0 0.03 4 0.1 3 0.2 (4,3) 3.5 0.02 4 0.1 4 0.1 (4,4) 4.0 0.01
Now we can find the Xbar for sample size n=2. Example: Find the probability of Xbar =3. This would be the sum of all the probability having a Xbar =3.
• .1 .03.04.03)4,2()3,3()2,4()3( +==+++== pppXP
TABLE 2: PMF for Xbar xbar 1 1.5 2 2.5 3 3.5 4 p(xbar) 0.16 0.24 0.25 0.2 0.1 0.04 0.01
Jeff Wetzel Engr 323 Problem 5.41
0
0.05
0.1
0.15
0.2
0.25
0.3
1 1.5 2 2.5 3 3.5 4 X bar
p(X bar)
Figure1: PMF of Xbar
Table 2 is the PMF of Xbar and Figure 1 is the graph of PMF for Xbar.
B) Looking at Table 2 we can now read the probability of Xbar2.5.
• .85 .2.25.24.16)5.2()5.2()2()5.1()1()5.2( +=+++=++=+ xxxxx pppppXP
C) We should use the hint they gave us and calculate the value of R for each outcome. R= the difference between the largest and smallest in the sample. In a similar way we created Table 1 we can now create a table with R values and their p(xbar). Table three calculates the probabilities of R=0,1,2,3 and Table 4 sums up the different probabilities that make the PMF of R. F
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