Question

Suppose that for a given computer salesperson, the probability distribution of x = the number of systems sold in 1 month is given by the following table.

x	1	2	3	4	5	6	7	8
p(x)	0.05	0.10	0.12	0.30	0.29	0.12	0.01	0.01

(a)

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

(b)

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

variance

standard deviation

How would you interpret these values? (Round your standard deviation to four decimal places.)

The mean squared deviation from the mean number of systems sold in one month is... . A typical deviation from the mean number of systems sold in one month is... .

(c)

What is the probability that the number of systems sold is within 1 standard deviation of its mean value?

(d)

What is the probability that the number of systems sold is more than 2 standard deviations from the mean?

Answer 1

a) The mean of value of X here is computed as:
Mean = Sum product of X and p(x)
Mean = 1*0.05 + 2*0.1 + 3*0.12 + 4*0.3 + 5*0.29 + 6*0.12 + 7*0.01 + 8*0.01
Mean = 4.13

Therefore 4.13 is the required mean here.

The second moment of X is computed here as:
E(X²) = 1²*0.05 + 2²*0.1 + 3²*0.12 + 4²*0.3 + 5²*0.29 + 6²*0.12 + 7²*0.01 + 8²*0.01 = 19.03

Therefore the variance here is computed as:
Var(X) = E(X²) - [E(X)]² = 19.03 - 4.13² = 1.9731
Therefore 1.9731 is the required variance here.

The standard deviation is computed now as:

Therefore 1.4047 is the required standard deviation here.

The mean squared deviation from the mean number of systems sold in one month is 1.9731 . A typical deviation from the mean number of systems sold in one month is 1.4047.

c) The probability that the number of systems sold is within 1 standard deviation of its mean value is computed here as:
P( Mean - SD(X) < X < Mean + SD(X) )

= P(4.13 - 1.4047 < X < 4.13 + 1.4047)
= P(2.73 < X < 5.53)
= P(X = 3) + P(X = 4) + P(X = 5)
= 0.12 + 0.3 + 0.29
= 0.71
Therefore 0.71 is the required probability here.

d) The probability that the number of systems sold is more than 2 standard deviations from the mean is computed here as:
P(X > Mean + 2SD(X))
= P(X > 4.13 + 2*1.4047)
= P(X > 6.9394)
= P(X = 7) + P(X = 8) = 0.02
Therefore 0.02 is the required probability here.