Question

1. number of credit cards a bank vice president feels that each savings account customer has, on average, three credit cards. the following distribution represents the number of credit cards people own. find the mean, variance, and standard deviation. is the vice president correct?

number of cards x | 0 1 2 3 4
probability p(x) | 0.18 0.44 0.27 0.08 0.03

Explanation:

Step1: Calculate the mean ($\mu$)

The formula for the mean of a discrete - probability distribution is $\mu=\sum x\cdot P(x)$.
\[

$$\begin{align*} \mu&=(0\times0.18)+(1\times0.44)+(2\times0.27)+(3\times0.08)+(4\times0.03)\\ &=0 + 0.44+0.54 + 0.24+0.12\\ &=1.34 \end{align*}$$

\]

Step2: Calculate the variance ($\sigma^{2}$)

The formula for the variance is $\sigma^{2}=\sum(x - \mu)^{2}\cdot P(x)$.
\[

$$\begin{align*} (0 - 1.34)^{2}\times0.18&=( - 1.34)^{2}\times0.18=1.7956\times0.18 = 0.323208\\ (1 - 1.34)^{2}\times0.44&=( - 0.34)^{2}\times0.44 = 0.1156\times0.44=0.050864\\ (2 - 1.34)^{2}\times0.27&=(0.66)^{2}\times0.27 = 0.4356\times0.27=0.117612\\ (3 - 1.34)^{2}\times0.08&=(1.66)^{2}\times0.08 = 2.7556\times0.08=0.220448\\ (4 - 1.34)^{2}\times0.03&=(2.66)^{2}\times0.03 = 7.0756\times0.03=0.212268 \end{align*}$$

\]
\[

$$\begin{align*} \sigma^{2}&=0.323208+0.050864 + 0.117612+0.220448+0.212268\\ &=0.9244 \end{align*}$$

\]

Step3: Calculate the standard deviation ($\sigma$)

The formula for the standard deviation is $\sigma=\sqrt{\sigma^{2}}$.
\[
\sigma=\sqrt{0.9244}\approx0.9615
\]

Answer:

The mean is $1.34$, the variance is $0.9244$, and the standard deviation is approximately $0.9615$. The vice - president is not correct since the mean number of credit cards is $1.34
eq3$.