Resistors labeled 100 have tue resistances that are between 80 and 120. Let X be the mass of randomly chosen resistor. The probability desnsity function of X is given by:

f(x)= {x-80/ 800, 80< x 120 and 0, otherwise

a. what proportion of resistors have resistances less than 90?
b. find the mean resistance.
c. find the standard deviation of the resistances.
d. find the cumulative distribution function of the resistances.

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 =

=0.0625

the probability of the resistors having  resistance  less than 90 is 0.0625

b) Mean=

  = 106.67

The mean resistance is

c. standard deviation of the resistances   ⇒

variance  

=88.89

standard deviation of the resistances =square root (88.89)

=9.42    

d. cumulative distribution function of the resistances.

=(0.5x^2-80x +3200)/800

17

step-by-step explanation:

51/3=17

thus 17 is your answer.

this "question" looks more like a project is there any other info you can give me about this. it could be too complex for someone else to do on this website