SOLUTION: A researcher calculated the values and probabilities for a random variable X as shown below. Unfortunately, he erased the last value and needs to figure out what it was. If the mea

Question 896413: A researcher calculated the values and probabilities for a random variable X as shown below. Unfortunately, he erased the last value and needs to figure out what it was. If the mean of X was 2.2, then what was the last value?
X
0
1
3
?
P(X)
0.4
0.1
0.2
0.3

Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A researcher calculated the values and probabilities for a random variable X as shown below. Unfortunately, he erased the last value and needs to figure out what it was. If the mean of X was 2.2, then what was the last value?
X
0
1
3
?
P(X)
0.4
0.1
0.2
0.3
------
Sounds like a weighted-mean problem.
Equation:
0.4*0 + 0.1*1 + 0.2*3 + 0.3*X = 2.2
------
0 + 0.1 + 0.6 + 0.3X = 2.2
0.7 + 0.3X = 2.2
0.3x = 1.5
X = 5
----------------------
Cheers,
Stan H.
Answer by jim_thompson5910(35256) (Show Source): You can put this solution on YOUR website!
let m = mean

m = 2.2 (given)

mean = Sum(value * probability)

m = 0*P(0) + 1*P(1) + 3*P(3) + x*P(x)

2.2 = 0*0.4 + 1*0.1 + 3*0.2 + x*0.3

2.2 = 0 + 0.1 + 0.6 + 0.3x

2.2 = 0.7 + 0.3x

2.2 - 0.7 = 0.3x

1.5 = 0.3x

0.3x = 1.5

x = 1.5/0.3

x = 5

So the missing value is 5.
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