Step 1

Given that, the X represents the number of packages being mailed by a randomly selected customer at a certain shipping facility.

The distribution of X is also provided as:

x	1	2	3	4
P(x)	0.4	0.3	0.2	0.1

 
Step 2

a.

Consider a random sample of size n = 2 (two customers), and let $\bar X$ be the sample mean number of packages shipped.

Required to obtain the probability distribution of $\bar X$.

Let the first observation be given by ${X_1}$ and second by ${X_2}$ respectively in a sample of 2 customers.

The distributions of ${X_1}$ and ${X_2}$ are independent as packages mailed by the customers are independent of one another.

Therefore, the joint probability distribution of pairs of sample observations $\left( {{x_1},{x_2}} \right)$ selected from ${X_1}$ and ${X_2}$ is given as:

\[p\left( {{x_1},{x_2}} \right) = {p_{{X_1}}}\left( {{x_1}} \right) \times {p_{{X_2}}}\left( {{x_2}} \right)\]

The joint pdf for the pair (1,1) is obtained as:

\[\begin{array}{c} p\left( {1,1} \right) = {p_{{X_1}}}\left( 1 \right) \times {p_{{X_2}}}\left( 1 \right)\\ = 0.4 \times 0.4\\ = 0.16 \end{array}\]

The joint pdf for all the other pairs $\left( {{x_1},{x_2}} \right)$ is similarly obtained as given below:

	x2					Total
x1	P(x1, x2)	1	2	3	4
	1	0.16	0.12	0.08	0.04	0.4
	2	0.12	0.09	0.06	0.03	0.3
	3	0.08	0.06	0.04	0.02	0.2
	4	0.04	0.03	0.02	0.01	0.1
	Total	0.4	0.3	0.2	0.1	1

let $\bar X$ be the sample mean number of packages shipped mailed by 2 customers.

The formula for calculating $\bar X$ is:

\[\bar X = \left( {\frac{{{x_1} + {x_2}}}{2}} \right)\]

The sample mean of packages shipped mailed by 2 customers for all pairs of observations $\left( {{x_1},{x_2}} \right)$ are given in the table below:

$\left( {{x_1},{x_2}} \right)$	$\bar X$	$p\left( {{x_1},{x_2}} \right)$
1,1	1	0.16
1,2	1.5	0.12
1,3	2	0.08
1,4	2.5	0.04
2,1	1.5	0.12
2,2	2	0.09
2,3	2.5	0.06
2,4	3	0.03
2,1	2	0.08
3,2	2.5	0.06
3,3	3	0.04
3,4	3.5	0.02
4,1	2.5	0.04
4,2	3	0.03
4,3	3.5	0.02
4,4	4	0.01

To obtain the distribution of $\bar X$, the values of $\bar X$ which are repeated are added with the unique values of $\bar X$ as shown:

$\bar x$	Probability	P($\bar x$)
1	0.16	0.16
1.5	0.12+0.12	0.24
2	0.08+0.09+0.08	0.25
2.5	0.04+0.06+0.06+0.04	0.20
3	0.03+0.04+0.03	0.10
3.5	0.02+0.02	0.04
4	0.01	0.01
Total	1	1

This is the required probability distribution of $\bar X$.

 
Step 3

b.

Refer to part (a), required to calculate $P\left( {\bar X \le 2.5} \right)$ .

Using probabilities from part (a), $P\left( {\bar X \le 2.5} \right)$ is calculated as shown:

\[\begin{array}{c} P\left( {\bar X \le 2.5} \right) = P\left( {\bar X = 1} \right) + P\left( {\bar X = 1.5} \right) + P\left( {\bar X = 2} \right) + P\left( {\bar X = 2.5} \right)\\ = 0.16 + 0.24 + 0.25 + 0.20\\ = 0.85 \end{array}\]

Therefore, the required probability $P\left( {\bar X \le 2.5} \right)$ is 0.85.

 
Step 4

c.

Again, consider a random sample of size n = 2, but now focus on the statistic R = the sample range (difference between the largest and smallest values in the sample).

Required to obtain the distribution of R.

Given that the statistic R = the sample range that is difference between the largest and smallest values in the sample.

That means, $R = \max \left\{ {{x_1},{x_2}} \right\} - \min \left\{ {{x_1},{x_2}} \right\}$

Following is the table which represents the probability corresponding to each pair $\left( {{x_1},{x_2}} \right)$ of R statistic.

(x1,x2)	R	P(x1,x2)
1,1	0	0.16
1,2	1	0.12
1,3	2	0.08
1,4	3	0.04
2,1	1	0.12
2,2	0	0.09
2,3	1	0.06
2,4	2	0.03
2,1	2	0.08
3,2	1	0.06
3,3	0	0.04
3,4	1	0.02
4,1	3	0.04
4,2	2	0.03
4,3	1	0.02
4,4	0	0.01

To obtain the distribution of R, the values of R which are repeated are added with the unique values of R as shown:

r	Probability	P(r)
0	0.16+0.09+0.04+0.01	0.3
1	0.12+0.12+0.06+0.06+0.02+0.02	0.4
2	0.08+0.03+0.08+0.03	0.22
3	0.04+0.04	0.08
Total	1	1

This is the required probability distribution of R.

 
Step 5

d.

If a random sample of size n = 4 is selected, required to find $P\left( {\bar X \le 1.5} \right)$.

Calculations:

$\bar X$ will be 1 when ${X_i} = 1$ for every i = 1,2,3,4.

$\bar X$ will be 1.25 when ${X_i} = 1$ in 3 out of 4 cases and ${X_i} = 2$ in 1 case.

The possibilities that $\bar X = 1.5$ are:

(a) When ${X_i} = 1$ in 2 out of 4 cases and ${X_i} = 2$ in remaining 2 cases.

(b) When ${X_i} = 1$ in 3 out of 4 cases and ${X_i} = 3$ in remaining 1 case.

As ${X_i}'s$ are independent, the probabilities are calculated as shown:

The probability $P\left( {\bar X = 1} \right)$ is obtained as,

\[\begin{array}{c} P\left( {\bar X = 1} \right) = P\left( {1,1,1,1} \right)\\ = {\left( {0.4} \right)^4}\\ = 0.0256 \end{array}\]

The probability $P\left( {\bar X = 1.25} \right)$ is obtained as,

\[\begin{array}{c} P\left( {\bar X = 1.25} \right) = \left( {\begin{array}{*{20}{c}} 4\\ 1 \end{array}} \right){\left( {0.4} \right)^3}{\left( {0.3} \right)^1}\\ = 0.0768 \end{array}\]

The probability $P\left( {\bar X = 1.5} \right)$ is obtained as,

\[\begin{array}{c} P\left( {\bar X = 1.5} \right) = \left( {\begin{array}{*{20}{c}} 4\\ 2 \end{array}} \right){\left( {0.4} \right)^2}{\left( {0.3} \right)^2} + \left( {\begin{array}{*{20}{c}} 4\\ 1 \end{array}} \right){\left( {0.4} \right)^3}{\left( {0.3} \right)^1}\\ = 0.0864 + 0.0512\\ = 0.1376 \end{array}\]

Now, the probability $P\left( {\bar X \le 1.5} \right)$ is obtained as:

\[\begin{array}{c} P\left( {\bar X \le 1.5} \right) = P\left( {\bar X = 1.5} \right) + P\left( {\bar X = 1.25} \right) + P\left( {\bar X = 1.5} \right)\\ = 0.0256 + 0.0768 + 0.1376\\ = 0.24 \end{array}\]

Therefore, the required probability $P\left( {\bar X \le 1.5} \right)$ is 0.24.