Mathematics: (a) A type of lightbulb is labeled as having an average lifetime of 1000 hours. It s reasonable to model the probability of failure of these bulbs by an exponential density function with mean μ = 1000. (i) Use this model to find the probability that a bulb fails within the first 300 hours. (Round your answer to three decimal places.) 0.259 Correct: Your answer is correct. (ii) Use this model to find the probability that a bulb burns for more than 900 hours. (Round your answer to three decimal places.)

A. of the 4 people who have a sister, 1 of them also has a brother.Step-by-step explanation:Hope it helped you!...

(a) A type of lightbulb is labeled as having an average

lexhoangg

21.01.2022

a)0.259

b)0.407

c)693.1hrs

Step-by-step explanation:

This is an exponential density function with a constant c and mean of 1000:

f(t)={0,t

The constant c is calculated as:

$\mu=\frac{1}{c}\\\\c=\frac{1}{\mu}=\frac{1}{1000}\\\\c=0.001$

#To find the probability that i falls with the first 300 hrs;

$P(0\leq X\leq 300)=\int\limits^{300}_0 {f(t)} \, dt\\\\=\int\limits^{300}_0{0.001e^{-0.001t}dt\\\\\\$

$=[-e^{-0.001t}]\limits^{300}_{0}\\\\=-0.7408+1\\\\=0.2592$

Hence, the probability that a bulb fails within the first 300 hours is 0.259

b. For more than 900hrs, integrate between 900 and infinity.

#Use the limit method for dealing with improper integrals:

$P(X\geq 900)=\int\limits^{\infty}_{900} {f(t)} \, dt\\ \\\\=\int\limits^{\infty}_{900}{0.001e^{-0.001t}dt\\\\$

$=\int\limits^{b}_{900}{0.001e^{-0.001t}dt, b-\infty\\\\$

$=[-e^{-0.001t}]\limits^{b}_{900}\\\\=[-e^{-0.001t}]\limits^{b}_{900}\\\\=0--0.4066\\\\=0.407$

Hence, the probability that a bulb burns for more than 900 hours is 0.407

c. To find the median lifetime of these bulbs:

$P(X\leq median)=P(X\geq median)=0.5\\\\P(X\geq median)=e^{-median/\mu}=0.5\\\\-median/1000=In(0.5)\\\\-median=1000\ In(0.5)\\\\median=693.1$

Hence, the median lifetime of these bulbs is 693.1hrs

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