Question

1. The time in minutes of labor required to complete a job at a bike shop is distributed according to a log-normal distribution with parameter values μ-3.8 and σ2-0.5 (these are not the same as the mean and variance of this random variable, see formulas). Assume that the times, X, required for the next 35 jobs are also independent so that X, LN(3.8,0.5) for j-1,.,35 is an lid collection of random variables. (a) Calculate the probability that the average number of minutes of labor required for the 35 jobs is less than 45 minutes

Answer 1

, the time required to complete a job at a bike shop has a log-normal distribution with parameters $\mu=3.8, \sigma^2=0.5$

The mean of X (that is expected value of X) is

0.5 = exp ( 3.84- = 57.3975

The variance of X is

7 3075

The standard deviation of X is

= V2137. 1915 = 46.2298 #x = V σ

a) Let $\begin{align*} \bar{X} \end{align*}$ be a random variable which is the sample mean of any given sample of size n=35 selected from X.

Using the Central Limit Theorem, since the sample size is greater than 30 (or we know the population standard deviation), we can say that $\begin{align*} \bar{X} \end{align*}$ has a normal distribution with

mean and

11 x = μ x = 57.3975

standard deviation (or standard error of mean)

46.2298_7 8143

The probability that the average number of minutes of labor required for 35 jobs is less than 45 minutes is same as the probability that $\begin{align*} \bar{X} \end{align*}$ < 45

P(X <45)- P get the z score of 45 45- 57.3975 7.8143 P(Z <-1.59) -P(Z > 1.59) Due to the symmetry of normal distribution PZ<-1.59)-P1.59) 1-P(Z < 1.59) = 1-(0.5 0.4441) using standard normal tables for z=1.59 0,0559

The probability that the average number of minutes of labor required for 35 jobs is less than 45 minutes is 0.0559