Answer:
a. The proportion of resistors having resistance less than 90 = 0.0625
b. Mean = 106.67
c. standard deviation of the resistances = 9.42
d. cumulative distribution function of the resistances
=(0.5x^2-80x +3200)/800
Step-by-step explanation:
Let X represent the mass of a randomly selected resistor
The probability density function given is
f(x)= x-80/800, 80<x<120
=0, otherwise
a) using the given pdf, the proportion of resistors having resistance less than 90 =
[tex]\int\limits^{90}_{80} {\frac{x-80}{800} } \, dx \\\\=\frac{1}{800} [0.5x^2- 80x]|^{90}_{80}\\\\=\frac{1}{800} [0.5(90)^2- 80(90)-(0.5(80)^2- 80(80))][/tex]
=0.0625
the probability of the resistors having resistance less than 90 is 0.0625
b) Mean=
[tex]\int\limits^{120}_{80} {xf(x)} \, dx = \int\limits^{120}_{80} {x\frac{x-80}{800} } \, dx\\\\= \frac{1}{800} \int\limits^{120}_{80} {x^2-80x} \, dx \\\\= \frac{1}{800} \times [\frac{x^3}{3} -\frac{80x^2}{2} ]|^{120}_{80}[/tex]
= 106.67
The mean resistance is
c. standard deviation of the resistances ⇒
variance
[tex]=\int\limits^{120}_{80} {(x-106.67)^2 f(x) } \, dx \\\\=\int\limits^{120}_{80} {(x-106.67)^2 \times \frac{x-80}{800} } \, dx[/tex]
=88.89
standard deviation of the resistances =square root (88.89)
=9.42
d. cumulative distribution function of the resistances.
[tex]\int\limits^x_{80} {f(x)} \, dx = \int\limits^x_{80} {\frac{x-80}{800} } \, dx \\\\=\frac{1}{800} [0.5 x^2 - 80x]|^x_{80}[/tex]
=(0.5x^2-80x +3200)/800
