MATH SOLVE

3 months ago

Q:
# Seven thousand lottery tickets are sold for $5 each. One ticket will win $2,000, two tickets will win $750 each, and five tickets will win $100 each. Let X denote the net gain from the purchase of a randomly selected ticket. a. Construct the probability distribution of X. b. Compute the expected value E(X) of X. Interpret its meaning. c. Compute the standard deviation σ of X.

Accepted Solution

A:

Answer:a) The distribution for the random variable X is given by: X | -5 | 95 | 745 | 1995 |P(X) | 6992/7000 | 5/7000 | 2/7000 | 1/7000 |b) E(X)=-4.43. That means if we buy an individual ticket by $5 on this lottery the expected value of loss if $4.43. c) [tex]Sd(X)=\sqrt{Var(X)}=\sqrt{738.947}=27.184[/tex]Step-by-step explanation:In statistics and probability analysis, the expected value "is calculated by multiplying each of the possible outcomes by the likelihood each outcome will occur and then summing all of those values".The variance of a random variable Var(X) is the expected value of the squared deviation from the mean of X, E(X).And the standard deviation of a random variable X is just the square root of the variance. Part aThe info given is:N=7000 represent the number of tickets sold$5 is the price for any ticketNumber of tickets with a prize of $2000 =1Number of tickets with a prize of $750=2Number of tickets with a prize of $100=5Let X represent the random variable net gain when we buy an individual ticket. The possible values that X can assume are:___________________________Ticket price Prize Net gain (X)___________________________5 2000 19955 750 745 5 100 955 0 -5___________________________Now we can find the probability for each value of XP(X=1995)=1/7000, since we ave just one prize of $2000P(X=745)=2/7000, since we have two prizes of $750P(X=95)=5/7000, since we have 5 prizes of $100P(X=-5)=6992/7000. since we have 6992 prizes of $0.So then the random variable is given by this tableX | -5 | 95 | 745 | 1995 |P(X) | 6992/7000 | 5/7000 | 2/7000 | 1/7000 |Part bIn order to calculate the expected value we can use the following formula:[tex]E(X)=\sum_{i=1}^n X_i P(X_i)[/tex]And if we use the values obtained we got:[tex]E(X)=(-5)*(\frac{6992}{7000})+(95)(\frac{5}{7000})+(745)(\frac{2}{7000})+(1995)(\frac{1}{7000})=\frac{-31000}{7000}=-4.43[/tex]That means if we buy an individual ticket by $5 on this lottery the expected value of loss if $4.43. Part cIn order to find the standard deviation we need to find first the second moment, given by :[tex]E(X^2)=\sum_{i=1}^n X^2_i P(X_i)[/tex]And using the formula we got:[tex]E(X^2)=(25)*(\frac{6992}{7000})+(9025)(\frac{5}{7000})+(555025)(\frac{2}{7000})+(3980025)(\frac{1}{7000})=\frac{5310000}{7000}=758.571[/tex]Then we can find the variance with the following formula:[tex]Var(X)=E(X^2)-[E(X)]^2 =758.571-(-4.43)^2 =738.947[/tex]And then the standard deviation would be given by:[tex]Sd(X)=\sqrt{Var(X)}=\sqrt{738.947}=27.184[/tex]