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The random variable X takes on the values of 2, 5, n and 15. The probability distribution of X is shown in the table below.

The expected value of X is 9.1. What is the value of n?

a) 8

b) 8.52

c) 10

d) 12

e) 14.4

Answer: d) 12

Step-by-step explanation: In a probability distribution, Expected Value is the average (or mean) value a distribution can assume.

The Expected Value, E(X), can be determine by:

[tex]E(X)=\Sigma xP(x)[/tex]

in which

x is how many times an event happens

P(x) is the probability of an event happening

The table below n is x and expected value is 9.1, then:

[tex]9.1=2*0.1+5*0.4+0.2n+15*0.3[/tex]

[tex]9.1=0.2n+6.7[/tex]

[tex]0.2n=2.4[/tex]

n = 12

Value of n in this probability distribution is 12.

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The value of n in the discrete probability distribution given is 12.

The the sum of [P(X) × X] in a discrete probability distribution is the expected value

From the discrete probability table given :

P(X) × X can be expressed as :

  • E(X), expected value = 9.1

(0.1 × 2) + (5 × 0.4) + (15×0.3) + 0.2n = 9.1

0.2 + 2.0 + 4.5 + 0.2n = 9.1

6.7 + 0.2n = 9.1

0.2n = 9.1 - 6.7

0.2n = 2.4

n = 2.4 / 0.2

n = 12

Therefore, the value of n in the table is 12

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