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 1.4. (round to one decimal place as needed.) the variance is. (round to one decimal place as needed.)

Explanation:

Step1: Recall variance formula

The formula for the variance $\sigma^{2}$ of a discrete - probability distribution is $\sigma^{2}=\sum(x - \mu)^{2}P(x)$, where $x$ are the values of the random variable, $\mu$ is the mean, and $P(x)$ is the probability of $x$. We know $\mu = 1.4$.

Step2: Calculate $(x-\mu)^{2}P(x)$ for each $x$

For $x = 0$: $(0 - 1.4)^{2}\times0.272=( - 1.4)^{2}\times0.272 = 1.96\times0.272=0.53312$.
For $x = 1$: $(1 - 1.4)^{2}\times0.293=( - 0.4)^{2}\times0.293 = 0.16\times0.293 = 0.04688$.
For $x = 2$: $(2 - 1.4)^{2}\times0.239=(0.6)^{2}\times0.239 = 0.36\times0.239 = 0.08604$.
For $x = 3$: $(3 - 1.4)^{2}\times0.137=(1.6)^{2}\times0.137 = 2.56\times0.137 = 0.35072$.
For $x = 4$: $(4 - 1.4)^{2}\times0.042=(2.6)^{2}\times0.042 = 6.76\times0.042 = 0.28392$.
For $x = 5$: $(5 - 1.4)^{2}\times0.017=(3.6)^{2}\times0.017 = 12.96\times0.017 = 0.22032$.

Step3: Sum up the values

$\sigma^{2}=0.53312 + 0.04688+0.08604 + 0.35072+0.28392+0.22032 = 1.521$.

Step4: Round the result

Rounding $1.521$ to one decimal place, we get $1.5$.

Answer:

$1.5$