Question

1. Calculate the mean, the variance, and the standard deviation of the following discrete probability distribution.

X	-23	17	-9	-3
P(X=x)	0.5	0.25	0.15	0.19

2. A marketing firm is considering making up to three new hires. Given its specific needs, the form feels that there is a 60% chance of hiring at least two candidates. There is only a 5% chance that it will not make any hires and a 10% chance that it will make all three hires.

*Find the expected value and standard deviation of the number of hires.

Answer 1

1.

mean = sum of (x*P(X))

= -23*0.5 + 17*0.25 + (-9)*0.15 + (-3)*0.19

= -9.17

mean = -9.17

variance = E(X^2) - E(X)^2

= sum of ((x^2)*P(X)) - ( sum of (x*P(X)) )^2

= ((-23)^2*0.5 + (17)^2*0.25 + (-9)^2*0.15 + (-3)^2*0.19) - mean^2

= 350.61 - (-9.17)^2

= 266.5211

variance = 266.5211

Standard deviation = variance^0.5 = 266.5211^0.5 = 16.3255

Standard deviation = 16.3255

2.

P(x>=2) = 0.60 {given}

P(2) + P(3) = 0.60

P(2) + 0.10 {given} = 0.60

P(2) = 0.50

P(0) = 0.05 {given}

P(3) = 0.10 {given}

P(0)+P(1)+P(2)+P(3) = 1

0.05+P(1)+0.50+0.10 = 1

P(1) = 0.35

E(number of hires(x)) = sum of (x*P(x))

= 0*0.05+1*0.35+2*0.50+3*0.10

= 1.65

expected value = 1.65

SD = sum of (x-E)^2 * P(x)

= (0-1.65)^2*0.05+(1-1.65)^2*0.35+(2-1.65)^2*0.50+(3-1.65)^2*0.10

= 0.5275

SD = 0.5275

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