Question

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 0.01
a.mean: 1.23; standard deviation: 0.44
b.mean: 1.30; standard deviation: 0.32
c.mean: 1.30; standard deviation: 0.44
d.mean: 1.23; standard deviation: 0.66

Answer 1

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)] = $\sum\limits^4_0 {x} \, P(x)$
                                       = 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) = $\sum\limits^4_0 {x^{2} } \, P(x)$
          = 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

Answer 2

The Mean and standard deviation of the given dataset come to be 1.23 and 0.66 respectively.

What is the standard deviation?

The standard deviation of a random variable, sample, statistical population, data set, or probability distribution is the square root of its variance.

x           P(x)       xP(x)                  x²               x² P(x)

0          0.07          0                     0                   0

1           0.68        0.68                  1                  0.68

2          0.21         0.42                  4                  0.84

3           0.03       0.09                  9                  0.27

4           0.01        0.04                  16                 0.16

ΣxP(x) = 1.23

So, Mean = 1.23

Σx² P(x) =1.95

(ΣxP(x))² = 1.23² = 1.513

So, Variance = Σx² P(x)  - (ΣxP(x))²

Variance = 1.95 - 1.513

Variance = 0.437

Standard deviation =√variance

Standard deviation =√0.437

Standard deviation = 0.66

Therefore, the Mean and standard deviation of the given dataset come to be 1.23 and 0.66 respectively.

To get more about standard deviation visit:

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