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 0.263 0.296 0.235 0.150 0.041 0.015 (a) find the mean, variance, and standard deviation of the probability distribution the me (round the var does not deviate from the mean. (round deviates from the mean by about 2 defects. the stan (round deviates from the mean by more than 2 defects. (b) inter deviates from the mean by about 1 defect. 1 or 2 defects. the standard deviation is 1.2, so the typical number of defects in a batch of 1000 (round to one decimal place as needed.)

Explanation:

Step1: Calculate the mean

The formula for the mean \(\mu\) of a discrete probability distribution is \(\mu=\sum_{x}x\cdot P(x)\).

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Step2: Calculate the variance

The formula for the variance \(\sigma^{2}=\sum_{x}(x - \mu)^{2}\cdot P(x)\)

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Step3: Calculate the standard deviation

The formula for the standard deviation \(\sigma=\sqrt{\sigma^{2}}\)

$$ \sigma=\sqrt{1.5}\approx1.2 $$

Answer:

The mean is \(1.5\), the variance is \(1.5\), and the standard deviation is \(1.2\)