Question

A particular brand of dishwasher soap is sold in three sizes: 25 oz, 40 oz, and 65 oz. Twenty percent of all purchasers select a 25-oz box, 50% select a 40-oz box, and the remaining 30% choose a 65-oz box. Let X1 and X2 denote the package sizes selected by two independently selected purchasers.
a. Determine the sampling distribution of... ),and compare to m.
b. Determine the sampling distribution of the sample variance S2, calculate E(S2), and compare to σ2.

Answer 1

Answer:

Step-by-step explanation:

Given that:

There are 3 sizes of dishwasher soaps.

Twenty percent of all purchasers select a 25-oz box, 50% select a 40-oz box, and the remaining 30% choose a 65-oz box. Let X1 and X2 denote the package sizes selected by two independently selected purchasers.

Therefore, $3 \times 3 = 9$ random samples of size 2 are taken from the population.

The information is well represented in the table below.

$x_1$     $x_2$     $p(x_1 \ x_2) = p(x_1)p(x_2)$    $\bar x = \dfrac{x_1+x_2}{2}$     $s^2= (x_1-\bar x)^2 +(x_2-\bar x)^2$

25    25    0.20×0.20=0.04        25                              0

25    40    0.20×0.50=0.10         32.5                         112.5

25    65    0.20×0.30= 0.06        45                            800

40    25    0.50×0.20=0.10         32.5                          112.5

40    40    0.50×0.50=0.25         40                             0

40    65    0.50×0.30=0.15          52.5                         312.5

65    25    0.30×0.20=0.06        45                             800

65    40    0.30×0.50=0.15          52.5                         312.5

65    65    0.30×0.30=0.09         65                              0

The sample distribution of $\bar x$ is shown below as ;

$\bar x$           25        32.5              40       45                  52.5          65

$p(\bar x)$       0.04   0.10+0.10      0.25   0.06+0.06    0.15+0.15    0.09    

                       = 0.20                      =0.12              =0.30

The value of $E(\bar x) = \sum \bar x \ P(\bar x)$

=  25(0.04) + 32.5(0.20) + 40(0.25) + 45(0.12)  +  52.5(0.30)  +  65(0.09)

= 44.5 oz

From the same table above ;

the sampling distribution of ${s^2}$ is be shown below;

${s^2}$          0                          112.5            312.5             800

           0.04+0.25       0.10+0.10     0.15+0.15        0.06+0.06

          +0.09

$p(s)^2$   = 0.38                =0.20           = 0.30               0.12

The value of $E(s)^2 =\sum s^2 .p(s)^2$

= (0×0.38)+(112.5×0.20)+(312.5×0.30)+(800×0.12)

=0+112+93.75+96

=301.75