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
Question
Answer:
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
solved
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11 months ago
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