Let X be the number of packages being mailed by a randomly selected customer at a certain shipping facility. Suppose the distribution of X is as follows. x | 1 2 3 4

p(x) | 0.4 0.3 0.2 0.1

Required:
a. Consider a random sample of size n=2 (two customers), and let Xbar be the sample mean number of packages shipped. Obtain the probability distribution of Xbar.
b. Use (a) to calculate P (Xbar <= 2.5).
c. If a random sample of size 4 is selected, what is P (Xbar <= 1.5)? (Hint: don't list all possible outcomes; only those for which xbar <= 1.5).
d. Find V(Xbar)

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Step-by-step explanation:

We first convert each mixed number to an improper fraction.

To change -6 7/8, we will ignore the negative at first.  Multiply the whole number by the denominator:

6(8) = 48

Add the numerator:

48+7 = 55

Putting the negative back in, we have -55/8.

To change -3 3/4, ignore the negative at first.  Multiply the whole number by the denominator:

3(4) = 12

Add the numerator:

12+3 = 15

Putting the negative back in, we have -15/4.

This gives us

-55/8 ÷ -15/4

To divide fractions, flip the second one and multiply:

-55/8 × -4/15

Multiply straight across:

(-55×-4)/(8×15) = 220/120

Simplifying this, we know that both 220 and 120 are divisible by 10; this gives us

22/12

Both 22 and 12 are even, so they are both divisible by 2.  This gives us

11/6

6 will go into 11 one time, with 5 remaining; this gives us

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