Problem 7 Suppose you have a random variable X that represents the lifetime of a
ID: 3375961 • Letter: P
Question
Problem 7 Suppose you have a random variable X that represents the lifetime of a certain brand of light ulbs. Assume that the lifetime of light bulbs are approximately normally distributed with mean 1400 and standard deviation 200 (in other words X~ N(1400, 200)). Answer the following using the standard normal distribution table: a) Approximate the probability of a light bulb lasting less than 1250 hours. b) Approximate the probability that a light bulb lasts between 1360 to 1460 hours. c) Approximate the probability that a light bulb lasts more than 1500 hours. d) Draw this distribution's shape. Note the value of X that is in the middle of the distribution and use the empirical rule to approximate the values that contain the middle 68% and 95% of all observations. e) Find the five number summary of light bulb lifetimes (approximate A- and A+ by finding the outlier criteria based on the z-scores of the quantiles). f) What is the probability that 225 randomly selected lightbulbs will have an average within 25 hours to the actual population mean?Explanation / Answer
Solution
Let X represent the life time (in hours) of the light bulb.
We are given X ~ N(1400, 2002)
Back-up Theory
If a random variable X ~ N(µ, ?2), i.e., X has Normal Distribution with mean µ and variance ?2, then, pdf of X, f(x) = {1/??(2?)}e^-[(1/2){(x - µ)/?}2] …………………………….(A)
Z = (X - µ)/? ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)
P(X ? or ? t) = P[{(X - µ)/?} ? or ? {(t - µ)/?}] = P[Z ? or ? {(t - µ)/?}] .………(2)
X bar ~ N(µ, ?2/n),…………………………………………………………….…….(3),
where X bar is average of a sample of size n from population of X.
So, P(X bar ? or ? t) = P[Z ? or ? {(?n)(t - µ)/? }] …………………………………(4)
Probability values for the Standard Normal Variable, Z, can be directly read off from
Standard Normal Tables or can be found using Excel Function ……………………..(5)
Part (a)
Probability of a light bulb lasting less than 1250 hours = P(X < 1250)
= P[Z < {(1250 - 1400)/200}] [vide (2)]
= P(Z < - 0.75)
= 0.2266 [vide (5)] ANSWER
Part (b)
Probability a light bulb lasts between 1360 and 1460 hours
= P(1360 < X < 1460)
= P[{(1360 - 1400)/200} < Z < {(1460 - 1400)/200}] [vide (2)]
= P(- 0.2 < Z < 0.3)
= P(Z < 0.3) - P(Z < - 0.2)
= 0.6179 – 0.4602 [vide (5)]
= 0.1577 ANSWER
Part (c)
Probability a light bulb lasts more than 1500 hours
= P(X > 1500)
= P[Z > {(1500 - 1400)/200}] [vide (2)]
= P(Z > 0.5)
= 0.3085 [vide (5)] ANSWER
Part (f)
Let Xbar = mean life of a randomly chosen sample of 225 light bulbs. Then, vide (3),
Xbar ~ N[1400, (40/3)2}
Probability the randomly chosen sample of 225 light bulbs will have average life within 25 hours of actual population mean = P[(1400 - 25) < Xbar < (1400 + 25)]
= P[-{25/(40/3)} < Z < {25/(40/3)}] [vide (2)]
= P(- 1.875 < Z < 1.875)
= 0.9392 ANSWER
DONE