Answer to Question #301986 in Statistics and Probability for toodles

Question

Question 2

The probability that a house in an urban area will be burglarized is 5%. A sample of 50 houses is randomly selected to determine the number of houses that were burglarized.

P (pass) = 0.05 q (failure) = 0.95 n = 50

(a)      Define the variable of interest, X.                                                      

(b)      What are the possible values of X?                                                    

(c)      What is the expected number of burglarized houses?                                  

(d)     What is the standard deviation of the number of burglarized houses?                      

(e)      What is the probability that none of the houses in the sample was burglarized?

Answer 1

a) we define X as the number of houses in urban areas that were burglarized.

There are 50 houses taken as a sample and the probability that has a house will be burglarized is 0.05, which we may define as the probability of success.

Thus x ~ Bin(50,0.05)

b) we define the pmf of x as p(X=x) = $\begin{pmatrix} 50 & \\ x& \end{pmatrix}$ (0.05)^x(0.95)^50-x x= 0,1,2 . . .. ,50

$0, otherwise$

so x can take values 0, 1, 2, 3, 4, 5, . . . . . ,50

c) E(x) = (50 * 0.05) = 2.5 which is approximately three houses.

d) standard deviation = (npq)^1/2

d\given p = 0.05, q= 1-p = 1-0.05 = 0.95

Thus standard deviation = (50 * 0.05 * 0.95)^1/2

= 1.541103501

e) we define p(x=0) = ( (50C0) * (0.05)⁰ * (0.95)⁵⁰ )

= 0.076944975 which is the required solution.