fifteen items or less: the number of customers in line at a supermarket express checkout counter is a random…

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.\nsend data to excel\n| x | p(x) |\n|----|----| \n| 0 | 0.05 |\n| 1 | 0.25 |\n| 2 | 0.30 |\n| 3 | 0.25 |\n| 4 | 0.10 |\n| 5 | 0.05 |\npart 1 of 3\n(a) compute the mean μx. round the answer to two decimal places.\nμx = 2.25\npart 2 of 3\n(b) compute the standard deviation σx. round your answer to 4 decimal places if necessary.\nσx = 1.2196\npart 3 of 3

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.\nsend data to excel\n| x | p(x) |\n|----|----| \n| 0 | 0.05 |\n| 1 | 0.25 |\n| 2 | 0.30 |\n| 3 | 0.25 |\n| 4 | 0.10 |\n| 5 | 0.05 |\npart 1 of 3\n(a) compute the mean μx. round the answer to two decimal places.\nμx = 2.25\npart 2 of 3\n(b) compute the standard deviation σx. round your answer to 4 decimal places if necessary.\nσx = 1.2196\npart 3 of 3

Answer

Explanation:

Step1: Recall mean formula

The mean $\mu_X$ of a discrete - random variable $X$ is given by $\mu_X=\sum_{x}x\cdot P(x)$. We have the following data: When $x = 0$, $P(0)=0.05$; when $x = 1$, $P(1)=0.25$; when $x = 2$, $P(2)=0.30$; when $x = 3$, $P(3)=0.25$; when $x = 4$, $P(4)=0.10$; when $x = 5$, $P(5)=0.05$. So, $\mu_X=(0\times0.05)+(1\times0.25)+(2\times0.30)+(3\times0.25)+(4\times0.10)+(5\times0.05)$.

Step2: Calculate each term

$0\times0.05 = 0$; $1\times0.25=0.25$; $2\times0.30 = 0.60$; $3\times0.25 = 0.75$; $4\times0.10=0.40$; $5\times0.05 = 0.25$.

Step3: Sum up the terms

$\mu_X=0 + 0.25+0.60 + 0.75+0.40+0.25=2.25$.

Step4: Recall standard - deviation formula

The standard deviation $\sigma_X$ of a discrete random variable $X$ is given by $\sigma_X=\sqrt{\sum_{x}(x - \mu_X)^2\cdot P(x)}$. 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 $x = 5$: $(5 - 2.25)^2\times0.05=(2.75)^2\times0.05 = 7.5625\times0.05=0.378125$.

Step5: Sum up the values

$\sum_{x}(x - \mu_X)^2\cdot P(x)=0.253125+0.390625+0.01875+0.140625+0.30625+0.378125 = 1.4875$.

Step6: Calculate the standard deviation

$\sigma_X=\sqrt{1.4875}\approx1.2196$.

Answer:

(a) $\mu_X = 2.25$ (b) $\sigma_X=1.2196$