Question

Suppose the probability density function of the length of computer cables is f left-parenthesis x right-parenthesis equals 0.1 from 1200 to 1210 millimeters. a) Determine the mean and standard deviation of the cable length. Mean = Entry field with correct answer 1205 millimeters Standard deviation = Entry field with correct answer 2.89 millimeters (Round the answer to 2 decimal places.) b) If the length specifications are 1192less-than x less-than 1204, what proportion of cables is within specifications?

Answer 1

Answer:

a) Mean = 1205

Standard Deviation = 2.89

b) P( 1192 < x < 1204) = 0.4

Step-by-step explanation:

We are given the following information in the question:

$f(x) = 0.1$

a = 1200, b = 1210

We are given a uniform distribution.

a) Mean:

$\mu = \displaystyle\frac{a+b}{2}\\\\\mu = \frac{1200+1210}{2} = 1205$

Standard Deviation:

$\sigma = \sqrt{\displaystyle\frac{(b-a)^2}{12}}\\\\= \sqrt{\displaystyle\frac{(1210-1200)^2}{12}} = \sqrt{8.33} = 2.89$

b) P( 1192 < x < 1204)

$=\displaystyle\int_{1192}^{1204} f(x) dx\\\\=\displaystyle\int_{1200}^{1204} (0.1) dx\\\\=0.1[x]_{1200}^{1204} = (0.1)(1204-1200) = 0.4$