A box contains 3 green and 2 red balls. Three balls are chosen one after the other. Determine the values of random variable G representing the number of green balls and compute for the mean and variance. Interpret the result. (Show solution)
Find a possible statistical problem inside your home relating to a discrete random variable lesson, explain in a paragraph form. (2-3 sentences) Then, find the possible random variable. Then create a probability distribution. Then compute for the mean, variance, and standard deviation. Then interpret result, and write conclusion. (Show solution)
1
Expert's answer
2021-05-31T16:36:38-0400
1.
p(G=1)=53⋅42⋅31=101
p(G=2)=53⋅42⋅32=51
p(G=3)=53⋅42⋅31=101
Expected value (the average number of green balls in the taken three balls):
E(X)=∑xipi=101+52+103=108=54
Variance (how far the number of taken green balls is spread out from their average value):
V(X)=E(X2)−(E(X))2
E(X2)=∑xi2pi=101+522+1032=1018=59
V(X)=59−5242=2529
2.
For example, there are five lamps in my flat. I have probabilities of the number of lamps which are switched on. And I need calculate the mean, variance and standard deviation.
Let n is the number of switched lamps.
Probability distribution:
p(n=0)=0.1
p(n=1)=0.2
p(n=2)=0.25
p(n=3)=0.15
p(n=4)=0.2
p(n=5)=0.1
Then:
the mean (the average number of switched lamps):
E(X)=0.2+2⋅0.25+3⋅0.15+4⋅0.2+5⋅0.1=2.45
Variance (how far the number of switched lamps is spread out from their average value):
V(X)=E(X2)−(E(X))2
E(X2)=0.2+4⋅0.25+9⋅0.15+16⋅0.2+25⋅0.1=8.25
V(X)=8.25−2.452=2.25
Standard deviation:
σ=V(X)=2.25=1.5
We can conclude, that there are three numbers of switched lamps (1,2,3) within one standard deviation (range is from 0.95 to 3.95).
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