Q:

The random variable x represents the number of credit cards that adults have along with the corresponding probabilities. find the mean and standard deviation. x p(x) 0 0.07 1 0.68 2 0.21 3 0.03 4 0.01a.mean: 1.23; standard deviation: 0.44b.mean: 1.30; standard deviation: 0.32c.mean: 1.30; standard deviation: 0.44d.mean: 1.23; standard deviation: 0.66

Accepted Solution

A:
Lets make a table first.

x                        P(x)
-----------------------------
0                         0.07
1                         0.68
2                         0.21 
3                         0.03
4                         0.01


Mean = Expected value[E(X)] = [tex] \sum\limits^4_0 {x} \, P(x)[/tex]
                                       = 0(0.07)+1(0.68)+2(0.21)+3(0.03)+4(0.01)
                                       = 1.23

Variance[V(X)] = E(X²) - [E(X)]²

E(X²) = [tex] \sum\limits^4_0 {x^{2} } \, P(x)[/tex]
          = 0²(0.07)+1²(0.68)+2²(0.21)+3²(0.03)+4²(0.01)
          =  1.95

Therefore, 
V(X) = E(X²)-[E(X)]² = 1.95-1.23² = 0.4371

Standard deviation(σ) = √V(X) = √0.4371 = 0.66