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Kitty [74]
3 years ago
6

Given the following probability distribution, what is the expected value of the random variable X? X P(X) 100 .10 150 .20 200 .3

0 250 .30 300 .10 Sum1.00 Multiple Choice 175 150 200 205
Mathematics
1 answer:
Mekhanik [1.2K]3 years ago
5 0

Answer:

d) 205

Step-by-step explanation:

<u><em>Step(i):-</em></u>

x             :   100      150      200      250      300

p(X=x)    :   0.10     0.20    0.30     0.30      0.10

<u>Step(ii):-</u>

Let 'X' be the discrete random variable

Expected value of the random variable

   E(X) = ∑ x P(X=x)

           =  100 X 0.10 + 150 X 0.20 + 200 X 0.30 +250 X 0.30 + 300 X 0.10

          =   205

<u>Final answer:</u>-

The expected value     E(X) = 205

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Step-by-step explanation:

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(x_2,y_2) = (7,1) --- Stadium

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This gives:

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Open bracket

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We have:

(x_1,y_1) = (3,4) --- Team hotel

(x_2,y_2) = (7,1) --- Stadium

The distance (d) is:

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1\ unit = 6.4\ miles

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3 years ago
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