Question

When parking a car in a downtown parking lot, drivers pay according to the number of hours or fraction thereof. The probability distribution of the number of hours cars are parked has been estimated as follows:

X	1	2	3	4	5	6	7	8
P(X)	0.224	0.142	0.106	0.08	0.057	0.039	0.033	0.319

A. Mean =

B. Standard Deviation =

The cost of parking is 2.25 dollars per hour. Calculate the mean and standard deviation of the amount of revenue each car generates.

A. Mean =

B. Standard Deviation =

Answer 1

SOLUTION:

From given data,

When parking a car in a downtown parking lot, drivers pay according to the number of hours or fraction thereof. The probability distribution of the number of hours cars are parked has been estimated as follows:

(A) Mean :

= (1*0.224) + (2*0.142)+ (3*0.106)+ (4*0.08)+ (5*0.057)+ (6*0.039)+ (7*0.033)+ (8*0.319)

= 4.448

E(X) =  4.448

= (1²*0.224) + (2²*0.142)+ (3²*0.106)+ (4²*0.08)+ (5²*0.057)+ (6²*0.039)+ (7²*0.033)+ (8²*0.319)

E(X²) = 27.888

var(X) =  E(X²) - [E(X)]²

= 27.888 - (4.448)²

= 27.888 - 19.784704

= 8.103296

(B) Standard deviation =

=

= 2.85

The cost of parking is 2.25 dollars per hour. Calculate the mean and standard deviation of the amount of revenue each car generates.

X	1	2	3	4	5	6	7	8
X * 2.25 =Y	1*2.25 = 2.25	2*2.25 =4.5	3*2.25 =6.75	4*2.25 =9	5*2.25 =11.25	6*2.25 =13.5	7*2.25 =15.75	8*2.25 =18
p(X)	0.224	0.142	0.106	0.08	0.057	0.039	0.033	0.319

(A) Mean:

= (2.25*0.224) + (4.5*0.142)+ (6.75*0.106)+ (9*0.08)+ (11.25*0.057)+ (13.5*0.039)+ (15.75*0.033)+ (18*0.319)

E(Y) = 10.008

=(2.25²*0.224) + (4.5²*0.142)+ (6.75²*0.106)+ (9²*0.08)+ (11.25²*0.057)+ (13.5²*0.039)+ (15.75²*0.033)+ (18²*0.319)

= 141.183

E(Y²) = 141.183

var(Y) =  E(Y²) - [E(Y)]²

= 141.183 - (10.008)²

= 141.183 - 100.160064

= 41.022936

(B) Standard deviation =

=

= 6.40