Question

overall assignment score: 85%
fifteen items or less: the number of customers in line at a supermarket express checkout counter is a random variable with the following probability distribution.
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x	p(x)
1	0.25
2	0.30
3	0.25
4	0.10
5	0.05

part 1 of 3
(a) compute the mean μ_x. round the answer to two decimal places.
μ_x = 2.25
part 2 of 3
(b) compute the standard deviation σ_x. round your answer to 4 decimal places if necessary.
σ_x = 1.2196
part 3 of 3
(c) if each customer takes 3 minutes to check out, what is the probability that it will take more than 6 minutes for all the customers currently in line to check out.

Explanation:

Step1: Recall mean formula for discrete - random variable

The formula for the mean $\mu_X$ of a discrete - random variable $X$ is $\mu_X=\sum_{x}x\cdot P(x)$.
We have:
\[

$$\begin{align*} \mu_X&=(0\times0.05)+(1\times0.25)+(2\times0.30)+(3\times0.25)+(4\times0.10)+(5\times0.05)\\ &=0 + 0.25+0.60 + 0.75+0.40+0.25\\ &=2.25 \end{align*}$$

\]

Step2: Recall variance formula for discrete - random variable

The formula for the variance $\sigma_{X}^{2}$ of a discrete - random variable $X$ is $\sigma_{X}^{2}=\sum_{x}(x - \mu_X)^{2}\cdot P(x)$.
First, calculate $(x-\mu_X)^{2}\cdot P(x)$ for each $x$:

• When $x = 0$: $(0 - 2.25)^{2}\times0.05=( - 2.25)^{2}\times0.05 = 5.0625\times0.05=0.253125$
• When $x = 1$: $(1 - 2.25)^{2}\times0.25=( - 1.25)^{2}\times0.25 = 1.5625\times0.25 = 0.390625$
• When $x = 2$: $(2 - 2.25)^{2}\times0.30=( - 0.25)^{2}\times0.30=0.0625\times0.30 = 0.01875$
• When $x = 3$: $(3 - 2.25)^{2}\times0.25=(0.75)^{2}\times0.25 = 0.5625\times0.25=0.140625$
• When $x = 4$: $(4 - 2.25)^{2}\times0.10=(1.75)^{2}\times0.10 = 3.0625\times0.10 = 0.30625$
• When $x = 5$: $(5 - 2.25)^{2}\times0.05=(2.75)^{2}\times0.05 = 7.5625\times0.05=0.378125$

Then $\sigma_{X}^{2}=0.253125 + 0.390625+0.01875+0.140625+0.30625+0.378125 = 1.4875$

Step3: Calculate standard deviation

The standard deviation $\sigma_X=\sqrt{\sigma_{X}^{2}}$.
$\sigma_X=\sqrt{1.4875}\approx1.2196$

Answer:

(a) $\mu_X = 2.25$
(b) $\sigma_X\approx1.2196$