SOLUTION: When a California household is randomly selected, the number of televisions and the corresponding probabilities are: 0(.03);1(.15);2(.29);3(.26);4(.16); 5(.11) Verify that this qu

Algebra ->  Probability-and-statistics -> SOLUTION: When a California household is randomly selected, the number of televisions and the corresponding probabilities are: 0(.03);1(.15);2(.29);3(.26);4(.16); 5(.11) Verify that this qu      Log On


   

Question 1070493: When a California household is randomly selected, the number of televisions and the corresponding probabilities are: 0(.03);1(.15);2(.29);3(.26);4(.16); 5(.11)
Verify that this qualifies as a probability distribution. Then find the average number of TV's per household and the standard deviation. Is it unusual for a household in California to not have a television?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
If it's a distribution,
0%3C=P%28x%29%3C=1
sum%28P%29=1
First one, check.
Second one,
.03%2B.15%2B.29%2B.26%2B.16%2B.11=1
Check.
.
.
.
mu=sum%28x%2AP%28x%29%29=0%2A.03%2B1%2A.15%2B2%2A.29%2B3%2A.26%2B4%2A.16%2B5%2A.11=2.7

sigma=1.29
So then a value of X=0 would be
2.7-N%2A1.29=0
N=2.1
This value is more than two standard deviations away from the mean.
Yes, it's unusual.