Given the probability distributions shown to the right, complete the following parts.

a. Compute the expected value for each distribution.

b. Compute the standard deviation for each distribution.

c. Compare the results of distributions A and B.

Distribution A: Distribution B:

X P(X) X P(x)

0 0.04 0 0.47

1 0.09 1 0.25

2 0.15 2 0.15

3 0.25 3 0.09

4 0.47 4 0.04

Round to three decimal places, compare the distributions for A and B.

a) Expected Value for distribution A, E(X) = 3.020

Expected Value for distribution B, E(X) = 0.980

b) Standard deviation of distribution A = 1.157

Standard deviation of distribution B = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

Step-by-step explanation:

Expected values is given as

E(X) = Σ xᵢpᵢ

where xᵢ = each possible sample space

pᵢ = P(X=xᵢ) = probability of each sample space occurring.

Distributions A and B is given by

X P(X) X P(x)

0 0.04 0 0.47

1 0.09 1 0.25

2 0.15 2 0.15

3 0.25 3 0.09

4 0.47 4 0.04

For distribution A

E(X) = Σ xᵢpᵢ = (0×0.04) + (1×0.09) + (2×0.15) + (3×0.25) + (4×0.47) = 3.02

For distribution B

E(X) = Σ xᵢpᵢ = (0×0.47) + (1×0.25) + (2×0.15) + (3×0.09) + (4×0.04) = 0.98

b) Standard deviation = √(variance)

But Variance is given by

Variance = Var(X) = Σx²p − μ²

where μ = E(X)

For distribution A

Σx²p = (0²×0.04) + (1²×0.09) + (2²×0.15) + (3²×0.25) + (4²×0.47) = 10.46

Variance = Var(X) = 10.46 - 3.02² = 1.3396

Standard deviation = √(1.3396) = 1.157

For distribution B

Σx²p = (0²×0.47) + (1²×0.25) + (2²×0.15) + (3²×0.09) + (4²×0.04) = 2.30

Variance = Var(X) = 2.30 - 0.98² = 1.3396

Standard deviation = √(1.3396) = 1.157

c) In distribution A, the bigger values of x have a higher probability of occurring than the values of distribution B (whose smaller values of x have a higher chance of occurring, hence, the expected value for distribution A is more than that of distribution B.

But according to the corresponding distribution of values, the two distributions have the same exact spread, A in ascending order (with higher values with bigger probability) and B in descending order (lower values have higher probabilities). But the same spread regardless, hence, the standard deviation which shows how data values spread around the mean (centre point) of a distribution is the same for the two distributions.

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl