Question

Suppose that for a given computer salesperson, the probability distribution of x = the number of systems sold in one month is given by the following table.

x	1	2	3	4	5	6	7	8
p(x)	0.06	0.10	0.11	0.30	0.31	0.10	0.01	0.01

(a) Find the mean value of x (the mean number of systems sold).
mean =  

(b) Find the variance and standard deviation of x. (Round the answers to four decimal places.)

variance	=
standard deviation	=

(c) What is the probability that the number of systems sold is within 1 standard deviation of its mean value?
P(μ − σ < x < μ + σ ) =  

(d) What is the probability that the number of systems sold is more than 2 standard deviations from the mean?
P(x < μ − 2σ or x > μ + 2σ) =

Answer 1

	x	p(x)	x*p(x)	x^2*p(x)
	1	0.06	0.06	0.06
	2	0.1	0.2	0.4
	3	0.11	0.33	0.99
	4	0.3	1.2	4.8
	5	0.31	1.55	7.75
	6	0.1	0.6	3.6
	7	0.01	0.07	0.49
	8	0.01	0.08	0.64
sum	36	1	4.09	18.73

(a) Find the mean value of x (the mean number of systems sold).
mean = sum(x*p(x))

= 4.09

(b) Find the variance and standard deviation of x. (Round the answers to four decimal places.)

var = sum(x^2*p(x)) - (sum(x*p(x))^2

var = 2.0019

standard deviation = sqrt(var) = 1.4148

(c) What is the probability that the number of systems sold is within 1 standard deviation of its mean value?
P(μ − σ < x < μ + σ ) = p(3<x<6) = 0.3+0.31 =0.61

(d) What is the probability that the number of systems sold is more than 2 standard deviations from the mean?
P(x < μ − 2σ or x > μ + 2σ) =p(x<1 or x>7) = p(x<1)+ p( x>7)

= 0 + 0.64

P(x < μ − 2σ or x > μ + 2σ) = 0.64