Question

The probability distribution of the number of accidents in North York, Ontario, each day is given by

x	0	1	2	3	4	5
P(x)	0.20	0.15	0.25	0.15	0.20	0.05

a) Based on this distribution, what would be the expected number of accidents on a given day?

A: 1.81

B: 1.47

C: 2.15

D: 4.62

b) Based on this distribution, what is the approximate value of the standard deviation of the number of accidents per day?

A: 2.15

B: 2.33

C: 6.95

D: 1.53

Answer 1

Solution:

x	P(x)	x * P(x)	x2 * P(x)
0	0.2	0	0
1	0.15	0.15	0.15
2	0.25	0.5	1
3	0.15	0.45	1.35
4	0.2	0.8	3.2
5	0.05	0.25	1.25
Sum	1	2.15	6.95

a)

Mean = X * P(X)

= 2.15

option c.

b)

Standard deviation =

=X 2 * P(X) - 2

=  6.95 - (2.15)2 = 2.3275

Standard deviation = = 1.53

option D.