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vikings1971 vikings1971
wrote...
Posts: 3
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9 years ago
The following data were obtained from a survey of college students. The variable X represents the number of non-assigned books read during the past six months.
 
x   0   1   2   3   4   5   6

P (X=x)   0.55   0.15   0.10   0.10   0.04   0.03   0.03
 
1. What is the standard deviation of X?  Place your answer, rounded to two decimal places in the blank. For example, 4.56 would be a legitimate entry.

2. Find P(X > 0). Place your answer, rounded to two decimal places in the blank. For example, 0.56 would be a legitimate entry.

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wrote...
9 years ago
Multiply each x value by its corresponding P(X = x) value

0*0.55 = 0
1*0.15 = 0.15
2*0.10 = 0.20
3*0.10 = 0.30
4*0.04 = 0.16

 
5*0.03 = 0.15
6*0.03 = 0.18

Now add up the products:
0+0.15+0.20+0.30+0.16+0.15+0.18 = 1.14

-------------------------------------------------------

So the expected value is 1.14
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