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

Question

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)

Answer 1

1.

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

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

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

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

$E(X)=\sum x_ip_i=\frac{1}{10}+\frac{2}{5}+\frac{3}{10}=\frac{8}{10}=\frac{4}{5}$

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

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

$E(X^2)=\sum x^2_ip_i=\frac{1}{10}+\frac{2^2}{5}+\frac{3^2}{10}=\frac{18}{10}=\frac{9}{5}$

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

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:

$p(n=0)=0.1$

$p(n=1)=0.2$

$p(n=2)=0.25$

$p(n=3)=0.15$

$p(n=4)=0.2$

$p(n=5)=0.1$

Then:

the mean (the average number of switched lamps):

$E(X)=0.2+2\cdot0.25+3\cdot0.15+4\cdot0.2+5\cdot0.1=2.45$

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

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

$E(X^2)=0.2+4\cdot0.25+9\cdot0.15+16\cdot0.2+25\cdot0.1=8.25$

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

Standard deviation:

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

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).