Answer to Question #200645 in Statistics and Probability for stress na si madam

Question #200645
  1. 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)
  2. 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)=352413=110p(G=1)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{1}{3}=\frac{1}{10}


p(G=2)=352423=15p(G=2)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{2}{3}=\frac{1}{5}


p(G=3)=352413=110p(G=3)=\frac{3}{5}\cdot\frac{2}{4}\cdot\frac{1}{3}=\frac{1}{10}


Expected value (the average number of green balls in the taken three balls):


E(X)=xipi=110+25+310=810=45E(X)=\sum x_ip_i=\frac{1}{10}+\frac{2}{5}+\frac{3}{10}=\frac{8}{10}=\frac{4}{5}


Variance (how far the number of taken green balls is spread out from their average value):


V(X)=E(X2)(E(X))2V(X)=E(X^2)-(E(X))^2


E(X2)=xi2pi=110+225+3210=1810=95E(X^2)=\sum x^2_ip_i=\frac{1}{10}+\frac{2^2}{5}+\frac{3^2}{10}=\frac{18}{10}=\frac{9}{5}


V(X)=954252=2925V(X)=\frac{9}{5}-\frac{4^2}{5^2}=\frac{29}{25}


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)=0.1p(n=0)=0.1

p(n=1)=0.2p(n=1)=0.2

p(n=2)=0.25p(n=2)=0.25

p(n=3)=0.15p(n=3)=0.15

p(n=4)=0.2p(n=4)=0.2

p(n=5)=0.1p(n=5)=0.1


Then:

the mean (the average number of switched lamps):

E(X)=0.2+20.25+30.15+40.2+50.1=2.45E(X)=0.2+2\cdot0.25+3\cdot0.15+4\cdot0.2+5\cdot0.1=2.45


Variance (how far the number of switched lamps is spread out from their average value):

V(X)=E(X2)(E(X))2V(X)=E(X^2)-(E(X))^2

E(X2)=0.2+40.25+90.15+160.2+250.1=8.25E(X^2)=0.2+4\cdot0.25+9\cdot0.15+16\cdot0.2+25\cdot0.1=8.25

V(X)=8.252.452=2.25V(X)=8.25-2.45^2=2.25


Standard deviation:

σ=V(X)=2.25=1.5\sigma=\sqrt{V(X)}=\sqrt{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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