Question

Your math professor receives several student emails each day. The probability model shows the number of emails your professor receives from students in a given day. Feel free to use StatCrunch or other technology to find your solutions. Enter only the solutions into the spaces provided. # student emails || 0 || 1 || 2 || 3 || 4 || 5 0.15 0.20 0.10 0.15 0.20 0.20 Probability 1. How many emails should your professor expect to receive daily? Round your result to the nearest hundredth. What is the standard deviation of the number of emails your professor should expect to receive? Round to the nearest hundredth 2 3. Iif it takes your professor an average of 5 minutes to reply to each email s/he recetves, how many should your professor expect to spend responding to student emails in a given day? 4. If your English professor receives the same mean and standard deviation number of emails, what is the mean and standard deviation of the difference in the number of emails your professors receive? Mean: Standard Deviation:

Answer 1

Let X = # Students emails

and p = Probability

Let;s use excel for calculation:

The formula used on the above excel sheet are as follows:

3) Expected time = 5* E(X) = 5 * 2.65 = 13.25

4)

Let Y = #Students mail recieved by English professor

So from the givven information E ( X ) = E( Y ) = 2.65

and SD ( X) = SD( Y )= 1.77

We want to find E(X - Y) = E( X ) - E ( Y ) = 2.65 - 2.65 = 0

and V(X - Y) = V( X) + V( Y) = 3.1275 + 3.1275 = 6.255

{here we assume that X and Y are independdent)

So that SD(X - Y) = $\sqrt{Var(X - Y)} = \sqrt{6.255} =\mathbf{ {\color{Red} 2.50}}$

Answer 2

SOLUTION :

1.

Expected student e-mails on a day, m :

= Sum (p * x) 

= 0.15 * 0 + 0.20 * 1 + 0.10 * 2 + 0.15 * 3 + 0.20 * 4+0.20 * 5

= 2 .65 student e-mails(ANSWER).

2.

Variance

= Sum (p * (x - m)^2 )

= 0.15*(0 - 2.65)^2+0.20*(1 - 2.65)^2+0.10*(2 - 2.65)^2++0.15*(3 - 2.65)^2+0.20*(4 - 2.65)^2+0.20*(5 - 2.65)^2

= 3.1275

SD 

= sqrt(variance) 

= sqrt(3.1275)

= 1.77 student e-mails (ANSWER)

3

Expected time spent in a day by the professor to respond to student e-mails

= Expected student e-mails in a day * time per e-mail

= 2.65 * 5

= 13.25 minutes (ANSWER)

4.

Mean(Math - English) = 0 ( all differences 0) 

=> Mean of difference = 0 (ANSWER).

Var (Math - English) = 0 (all differences 0)

=> SD = 0  (ANSWER)