TU Wien:Statistik und Wahrscheinlichkeitstheorie UE (Levajkovic)/Übungen 2023W/HW04.5

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 she sells each day is normally distributed with an expectation of 320 cups and a standard deviation of 20 cups. She also believes that the number of doughnuts she 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 she will sell more than 1900 cups of coffee in a week?

(b) If she makes a profit of 1.2 euro on each cup of coffee and 0.7 euro on each doughnut, can she reasonably expect to have a day’s profit over 550 euro? Justify your answer.

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Hilfreiches

Ähnliches Beispiel: 2019W - HW4.4

Lösung von Tutorin

(a)

Denote by $X_{i}$ the number of cups of coffee sold on $i$th day of the week $i=1,\dots ,6$. It is assumed $X_{i}\sim {\mathcal {N}}(320,20^{2})$ and $X_{i}$ and $X_{j}$, for $i\neq j$ are indepentent.

Let $X=X_{1}+\dots +X_{6}$. The goal is to calculate $P(X>1900)$. First, $X$ is the sum of indepentent normally distributed random variables and thus it is also normally distributed with

$$\mathbb {E} X=\mathbb {E} (X_{1}+\dots +X_{6})=\mathbb {E} (X_{1})+\dots +\mathbb {E} (X_{6})=6\cdot 320=1920$$

$$\mathbb {V} arX=\mathbb {V} ar(X_{1}+\dots +X_{6})\ {\overset {\underset {\mathrm {indep.} }{}}{=}}\ \mathbb {V} ar(X_{1})+\dots +\mathbb {V} ar(X_{6})=6\cdot 20^{2}=2400.$$

Thus, $X\sim {\mathcal {N}}(1920,2400)$ and

$$Z={\frac {X-1920}{\sqrt {2400}}}\sim {\mathcal {N}}(0,1).$$

The probability the shop owner will sell more than 1900 cups of coffee in a week equals

$${\begin{aligned}P(X>1900)&=1-P(Z\leq {\frac {1900-1920}{\sqrt {2400}}}\\&=1-\Phi \left({\frac {1}{\sqrt {6}}}\right)\\&=1-{\texttt {pnorm(1/sqrt(6))}}\approx 0.6584543.\end{aligned}}$$

Or directly

$$P(X>1900)={\texttt {pnorm(1900,1920,sqrt(2400),lower.tail=FALSE)}}\approx 0.6584543.$$

pnorm(1900, 1920, sqrt(2400), lower.tail=FALSE)
[1] 0.6584543

# or
1-pnorm(1900, 1920, sqrt(2400))
[1] 0.6584543

(b)

We denote by $Y_{i}$ the number of doughnuts sold on the $i$th day, $i=1,\dots ,6$. It is assumed that $Y_{i}\sim {\mathcal {N}}(150,12^{2})$. It is also assumed that $X_{i}$ and $Y_{i}$ are independent. Let $P_{i}=1.2X_{i}+0.7Y_{i}$ be the profit on the $i$th day, $i=1,\dots ,6$. Then, $P_{i}$ are normally distributed with

$$\mathbb {E} (P_{i})=\mathbb {E} (1.2X_{i}+0.7Y_{i})=1.2\cdot \mathbb {E} (X_{i})+0.7\cdot \mathbb {E} (Y_{i})=1.2\cdot 320+0.7\cdot 150=489$$

$$\mathbb {V} arP_{i}=\mathbb {V} ar(1.2X_{i}+0.7Y_{i})\ {\overset {\underset {\mathrm {indep.} }{}}{=}}\ 1.2^{2}\cdot \mathbb {V} ar(X_{i})+0.7^{2}\cdot \mathbb {V} ar(Y_{i})=1.2^{2}\cdot 20^{2}+0.7^{2}\cdot 12^{2}=646.56$$

Then,

$$Z={\frac {P_{i}-489}{\sqrt {646.56}}}\sim {\mathcal {N}}(0,1)$$

and

$$P(P_{i}\geq 550)=1-P\left({\frac {550-489}{\sqrt {646.56}}}\right)=1-\Phi (2.398973)\approx 0.00822$$

or directly

$$P(P_{i}>550)={\texttt {pnorm(550,489,sqrt(646.56),lower.tail=FALSE)}}\approx 0.00822.$$

There is a probability of 0.8% for a daily profit over 550 Euro.