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Talja [164]
3 years ago
15

The range of the random variable X is {1, 10, 15, 20, 25, x} where x is unknown. The value x has probability 0.50 and each of th

e remaining values has probability 0.10. The expected value of X is 22.1. Determine x. Answer to 1 decimal places.
Mathematics
1 answer:
marin [14]3 years ago
6 0

Answer:

22.1 = 1*0.1 +10*0.1 +15*0.1 + 20*0.1 +25*0.1 + x*0.5

And we can solve for the value of x and we got:

22.1 -[1*0.1 +10*0.1 +15*0.1 + 20*0.1 +25*0.1 ] = 0.5 x

15 = 0.5 x

x =30.0

Step-by-step explanation:

For this case we have the following distribution:

X        1       10     15    20   25    x

P(X)   0.1    0.1    0.1   0.1   0.1   0.5

The expected value of a random variable X is the n-th moment about zero of a probability density function f(x) if X is continuous, or the weighted average for a discrete probability distribution, if X is discrete.

The variance of a random variable X represent the spread of the possible values of the variable. The variance of X is written as Var(X).  

The expected value is given by:

E(X) = \sum_{i=1}^n X_i P(X_i)

And replacing we have this:

22.1 = 1*0.1 +10*0.1 +15*0.1 + 20*0.1 +25*0.1 + x*0.5

And we can solve for the value of x and we got:

22.1 -[1*0.1 +10*0.1 +15*0.1 + 20*0.1 +25*0.1 ] = 0.5 x

15 = 0.5 x

x =30.0

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