Question

use the probability distribution to complete parts (a) and (b) below. the number of defects per 1000 machine parts inspected defects 0 1 2 3 4 5 probability 0.272 0.293 0.239 0.137 0.042 0.017 (a) find the mean, variance, and standard deviation of the probability distribution. the mean is . (round to one decimal place as needed.)

Explanation:

Step1: Recall mean formula

The mean $\mu$ of a discrete - probability distribution is given by $\mu=\sum_{x}x\cdot P(x)$, where $x$ is the value of the random variable and $P(x)$ is its probability.
\[

$$\begin{align*} \mu&=(0\times0.272)+(1\times0.293)+(2\times0.239)+(3\times0.137)+(4\times0.042)+(5\times0.017)\\ &=0 + 0.293+0.478 + 0.411+0.168+0.085\\ &=1.435 \end{align*}$$

\]

Step2: Recall variance formula

The variance $\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)$.
First, calculate $(x-\mu)^{2}\cdot P(x)$ for each $x$:

• For $x = 0$: $(0 - 1.435)^{2}\times0.272=( - 1.435)^{2}\times0.272 = 2.059225\times0.272\approx0.560$
• For $x = 1$: $(1 - 1.435)^{2}\times0.293=( - 0.435)^{2}\times0.293 = 0.189225\times0.293\approx0.055$
• For $x = 2$: $(2 - 1.435)^{2}\times0.239=(0.565)^{2}\times0.239 = 0.319225\times0.239\approx0.076$
• For $x = 3$: $(3 - 1.435)^{2}\times0.137=(1.565)^{2}\times0.137 = 2.449225\times0.137\approx0.335$
• For $x = 4$: $(4 - 1.435)^{2}\times0.042=(2.565)^{2}\times0.042 = 6.589225\times0.042\approx0.277$
• For $x = 5$: $(5 - 1.435)^{2}\times0.017=(3.565)^{2}\times0.017 = 12.709225\times0.017\approx0.216$

Then $\sigma^{2}=0.560 + 0.055+0.076+0.335+0.277+0.216 = 1.529$

Step3: Recall standard - deviation formula

The standard deviation $\sigma=\sqrt{\sigma^{2}}$. So $\sigma=\sqrt{1.529}\approx1.24$

Answer:

The mean is $1.4$