Answer to Question #330637 in Statistics and Probability for kylie

Question

You pay 168 dollars to play a lottery. A bin contains 91 balls labeled from 1 to 91. You draw a ball and you receive cash in the amount of 3 times the number shown on the ball. Let X be the number shown on the ball.

(a) Identify the probability distribution for the X.

Distribution = Parameter =    

(b) Calculate the probability that you get more money back than you have spent on the ticket.

P(you win) =    

(c) Calculate the average number that you will draw. Calculate the standard deviation for this number.

μ =    

σ =    

(d) Calculate the average amount that you will gain (winnings minus what you paid) by purchasing a ticket. Calculate the standard deviation associated to this number.

Average Gain (Winnings - paid) =    

SD of Gain =    

Answer 1

a) We can select each ball with probability of $1/91\approx0.011$

$P(X)=0.011~for~X~from~1~to~91$

This is a uniform distribution.

$b)~P(winning)=P(3X>168)=P(X>56)=\\ =(91-56)/91=5/13\approx0.385$

$c)~\mu=\frac{1}{91}\sum_{k=1}^{91}k=\frac{1}{91}\frac{91(91+1)}{2}=46\\ \sigma^2=\frac{1}{90}\sum_{k=1}^{91}(k-\mu)^2=\frac{1}{90}(\sum_{k=1}^{45}(46-k)^2+\\ +\sum_{k=47}^{91}(k-46)^2)=\frac{1}{90}(\sum_{k=1}^{45}k^2+\sum_{k=1}^{45}k^2)=\\ =\frac{1}{45}\sum_{k=1}^{45}k^2$

For the sum of squares we will use the following formula:

$\sum_{k=1}^nk^2=\frac{1}{6}n(n+1)(2n+1)$

$\sigma^2=\frac{1}{45\cdot6}45\cdot46\cdot91\approx698\\ \sigma=\sqrt{\sigma^2}\approx26.4$

d) Average gain:

$G=\frac{1}{91}\sum_{k=1}^{91}(3k-168)=\\ =3\mu-168=-30$

Variance of gain:

$SD^2=\frac{1}{90}\sum_{k=1}^{91}((3k-168)-(-30))^2=\\ =\frac{1}{90}\sum_{k=1}^{91}(3k-138)^2=\frac{1}{90}(9\sum_{k=1}^{91}k^2-\\ -2\cdot3\cdot138\sum_{k=1}^{91}k+91\cdot138^2)=\\ =\frac{1}{90}(9\cdot\frac{1}{6}\cdot91\cdot92\cdot183-138(6\cdot\frac{91\cdot92}{2}-\\ -91\cdot138))=6279$

Standard deviation of gain:

$SD=\sqrt{SD^2}\approx79.24$