Ann has 30 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. Bob has 30 jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. Ann's and Bob's times are:

Assuming this problem: "Ann has 30 jobs that she must do in sequence, with the times required to do each of these jobs being independent random variables with mean 50 minutes and standard deviation 10 minutes. Bob has 30 jobs that he must do in sequence, with the times required to do each of these jobs being independent random variables with mean 52 minutes and standard deviation 15 minutes. Ann's and Bob's times are independent. Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs".

Previous concepts

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

For this case we can create some notation.

Let A the values for Ann we know that n1 = 30 jobs solved in sequence and we can assume that the random variable [tex]X_i[/tex] the time in order to do the ith job for [tex]i=1,2,....,n_1[/tex]. We will have the following parameters for A.

[tex]\mu_A = 50, \sigma_A =10[/tex]

W can assume that B represent Bob we know that n2 = 30 jobs solved in sequence and we can assume that the random variable [tex[X_i[/tex] the time in order to do the ith job for [tex]i=1,2,....,n_2[/tex]. We will have the following parameters for A

[tex]\mu_B = 52, \sigma_B =15[/tex]

And we can find the distribution for the total, if we remember the definition of mean we have:

[tex]\bar X= \frac{\sum_{i=1}^n X_i}{n}[/tex]

And [tex]T =n \bar X[/tex]

And the [tex]E(T) = n \mu[/tex]

[tex]Var(T) = n^2 \frac{\sigma^2}{n}=n\sigma^2 [/tex]

So then we have:

[tex]E(T_A)=30*50 =1500 , Var(T_A) = 30*10^2 =3000[/tex]

[tex]E(T_B)=30*52 =1560 , Var(T_B) = 30 *15^2 =6750[/tex]

Since we want this probability "Find the approximate probability that Ann finishes her jobs before Bob finishes his jobs" we can express like this:

[tex]P(T_A < T_B) = P(T_A -T_B<0)[/tex]

Since we have independence (condition given by the problem) we can find the parameters for the random variable [tex]T_A -T_B [/tex]

[tex]E[T_A -T_B] = E(T_A) -E(T_B)=1500-1560=-60[/tex]

[tex]Var[T_A -T_B]= Var(T_A)+Var(T_B) =3000+6750=9750[/tex]

And now we can find the probability like this:

[tex]P(T_A < T_B) = P(T_A -T_B<0)[/tex]

[tex]P(\frac{(T_A -T_B)-(-60)}{\sqrt{9750}}< \frac{60}{\sqrt{9750}})[/tex]

[tex]P(Z<0.608) =0.728[/tex]