Which two expressions are equivalent?

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

Elan Coil [88]

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.

7 0

3 years ago

Can someone help me find the standard deviation of this data set plssss

Svetllana [295]

since it was too long, I just made this python code in order to solve for it:

from math import *

data = [35, 27, 24, 42, 25, 31, 41, 34, 32, 46, 29, 44, 29, 35, 39, 27, 41, 63, 52, 31, 38, 46, 39, 27, 58, 32, 33, 47, 36, 28, 65, 53, 40, 32, 39, 44, 48, 59, 46, 62, 22, 39, 35, 29, 36, 66, 33, 45, 50, 62, 27, 27, 41, 33, 54, 44, 26, 31, 45, 33, 42, 63, 51, 70, 42, 53, 38, 28, 35, 49, 47, 54, 44, 40, 39, 62, 30, 35, 38, 26, 34, 61, 19, 30, 34, 26, 41, 35, 26, 54, 28, 32, 50, 27, 31, 35, 38, 26, 47, 30, 28, 62, 57, 56, 30, 32, 31, 36, 32, 42, 33, 51, 35, 30, 28, 20, 23, 27, 33, 36, 35, 42, 43, 28, 36, 44, 58, 41, 36, 37, 49, 34, 23, 39, 34, 37, 62, 23, 28, 26, 29, 30, 37, 46, 43, 56, 53, 31, 36, 44, 39, 51, 21, 51, 34, 30, 25, 31, 38, 52, 49, 43, 50, 31, 36, 32, 28, 44, 61, 57, 24, 63, 28, 31, 32, 26, 28]

def stand_dev(list_n):

x = 0

sum_num = 0

for x in list_n:

sum_num += list_n[x]

sum_of_n = sum_num

mean = float(sum_of_n / len(list_n))

float(mean)

i = 0

e = 0.0

for i in list_n:

subtract_then_square = (list_n[i] - mean)**2

e += subtract_then_square

result1 = e

new_mean = result1 / len(list_n)

final_result = float(sqrt(new_mean))

return final_result

print(stand_dev(data))

basically, the first chunk of the code calculated the mean of the data. The second chunk would take all of the values from the data, subtract it from the mean, and then square it. I then took the sum of all the new values and average it to get the new mean, then square rooted the new mean to get the answer.

it should be ~= 11.713, but you can always check it by searching up a python compiler online and then insert in the data.

7 0

3 years ago

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Ne4ueva [31]

The value is -58.5
hope it helps

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