The probability density function of the time to failure of an electronic component in a copier (in hours) is f(x) = (e^-x/1076)/(1076) for x>0. Determine the probability that

(A) A component lasts more than 3000 hours before failure. (Round the answer to 3 decimal places.)

(B) A component fails in the interval from 1000 to 2000 hours. (Round the answer to 3 decimal places.)

(C) A component fails before 1000 hours. (Round the answer to 3 decimal places.)

a. P(X>3000) = 0.062

b. P(1000 > X > 2000) = 0.239

c. P(X<1000) = 0.605

Explanation:

Given

The General Probability Density Function of the random variable is is given as follows:

f(x) = e^-(x/1076)/1076 for x >0

To calculate the probability, we make use of

P(a<X<b) = 1/1076 ∫ e^-(x/1076) dx {b,a}

P(a<X<b) = e^-(x/1076) {b,a}

P(a<X<b) = e^-(a/1076) - e^-(b/1076)

a. P(X>3000)

This translates to

P(∞ > X > 3000)

= e^-(3000/1076) - e^-(∞/1076)

= e^-(3000/1076) - 0

= e^-(3000/1076)

= 0.061537773515010

= 0.062

b.

P(1000 > X > 2000)

= e^-(1000/1076) - e^-(2000/1076)

= 0.238933619675996

= 0.239

c.P(X<1000)

This translates to

P(0 > X > 1000)

= e^-(0/1076) - e^-(1000/1076)

= 0.605196864611088

= 0.605

b. check your rearview mirror

explanation:

answer:

so whats the question ill edit my answer once ik

Explanation: