A. For a group of individuals, the random variable x denotes the number of credit cards per individual with the following distribution. x: 0 1 2 3 4 5 P(x): .27 .28 .20 .15 .08 .02 a. find the mean, variance and standard deviation of x b. find the probability that a randomly selected individual holds at least 1 card.

1.55

2.66

1.631

Step-by-step explanation:

Given :

x: 0 1 2 3 4 5

P(x): .27 .28 .20 .15 .08 .02

The expected mean, E(X) = Σ(x*p(x))

E(X) = (0*0.27)+(1*0.28)+(2*0.20)+(3*0.15)+(4*0.08)+(5*0.02)

E(X) = 1.55

The expected variance :

Σx²*p(x) - E(X)

(0^2*0.27)+(1^2*0.28)+(2^2*0.20)+(3^2*0.15)+(4^2*0.08)+(5^2*0.02) - 1.55

4.21 - 1.55 = 2.66

The standard deviation :

√variance = √2.66 = 1.631

5/6

step-by-step explanation:

if i remember correctly with perpendicular stuff you just take away the negative

range is infinite on both x and y axis positive and negative

step-by-step explanation: