Given the probability distributions shown to the​ right, complete the following parts. a. Compute the expected value for each d

Question

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.

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Camila 4 days 2021-10-09T22:01:29+00:00 1 Answer 0

Answers ( )

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    2021-10-09T22:02:49+00:00

    Answer:

    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.

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