TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Bura)/Übungen 2019W/4.4

Coffee and doughnuts

At a certain coffee shop, all the customers can buy a cup of coffee and also a doughnut. The shop owner believes that the number of cups he sells each day is normally distributed with an expectation of 320 cups and a standard deviation of 20 cups. He also believes that the number of doughnuts he sells each day is independent of the coffee sales and is normally distributed with an expectation of 150 doughnuts and a standard deviation of 12.

(a) The shop is open every day but Sunday. Assuming day-to-day sales are independent, what is the probability he will sell more than 2000 cups of coffee in a week?

(b) If he makes a profit of 50 cents on each cup of coffee and 40 cents on each doughnut, can he reasonably expect to have a day’s profit over 300 euro? Justify your answer.

Lösungsvorschlag von Gittenburg

--Gittenburg 18:06, 3. Nov. 2019 (CET)

(a)

$Var(6x)=6Var(x)=6*20^{2}=2400$

$\sigma ={\sqrt {2400}}=48.99$

> 1 - pnorm(2000, 6*320, 48.99)
[1] 0.05123594

(b)

Expected profit: $320\cdot 0.5+150\cdot 0.4=220$

$Var(0.5C)+Var(0.4D)=0.25\cdot 20^{2}+0.16*12^{2}=123.04$

$\sigma ={\sqrt {123.04}}=11.09$

No, 300€ is more than 7 SD away from the mean.

Lösungsvorschlag von Kave

--Kave 16:44, 27. Feb. 2020 (CET)

(a)

$z={\frac {x-\mu }{\sigma }}={\frac {2000-6(320)}{6(20)}}\approx 0.67$

from z-table 0.67 => 0.7486

$P(X>2000)=1-0.7486=0.2514=25.14\%$

Vervollständigen/Lösungsvorschlag von Jeremyer

--Jeremyer 16:44, 27. Feb. 2020 (CET)

(a) Kaves Idea is correct but there are some mistakes

For the SD of week, we need to do the following

$sd1w={\sqrt {var(1day)*6}}={\sqrt {20^{2}*6}}=48.9897948557$

Then we can just plug it into the formular and look it up in the table

$z={\frac {x-\mu }{\sigma }}={\frac {2000-6(320)}{\sqrt {20^{2}*6}}}\approx 1.63$

from z-table 1.63 => 0.9484

$P(X>2000)=1-0.9484=0.0516=5.16\%$

Which is also approximatly the same value as Gittenburg

(b) R-Studio same values as Solution from Gittenburg

Calculate expected profit

> profit = 320*0.5 + 150*0.4

Calculate standard deviation

1. standard deviation for the income is the sqrt of the sum of the deviation of
2. var coffees * profit coffee + var doughnuts * profit doughnuts

> var = (20*0.50)^2 + (12*0.40)^2
> sd = sqrt(var)

1. Plug it into the formular

> 1-pnorm(300, profit, sd)
[1] 2.753353e-13

Can one expect to have a day's profit over 300?

-> No, because the wanted profit is over 7 sd away from mean, this means it is very unlikely