Question

Question 2: Discrete Random Variable

 

The BRIT Sports Bar sells a large quantity of Guinness every Saturday. From past records, the pub has determined the following probabilities for sales:

 

 

Number of Crates (X) 

P(x)

18-0.15

19-0.10

20-0.32

21-0.05

22-0.13

23-0.25

 

a. Verify that this [P(x)] is a probability distribution. ​(2 marks)

b. Find the probability that the number of crates sold will be at least 22.  (3 marks)

c. Find the probability that the number of crates sold will be at most 20.  (3 marks)

d. What is the expected value for the number of crates sold for any given Saturday? 

(4 marks)

 

e. Calculate the variance of the distribution? ​(6 marks)

f. Determine the standard deviation of the distribution? ​(2 marks)

can workout be shown please.

Answer 1

a. To verify P(X) is a probability function we have to show that $\sum_{X} P(X) = 1$.

$\begin{aligned} \sum_{x} P(x) &= P(18) + P(19) + P(20) + P(21)+P(22)+P(23)\\ &= 0.15+0.10+0.32+0.05+0.13+0.25\\ &=1 \end{aligned}$

Hence P(X) is a probability function.

b.

$\begin{aligned} P(\text{number of crates atleast 22}) &= P(X \ge 22)\\ &= P(X = 22) + P(X = 23)\\ &= 0.13+0.25 = 0.38 \end{aligned}$

c.

$\begin{aligned} P(\text{number of crates atmost 20}) &= P(X \le 20)\\ &= P(X = 18) + P(X = 19) + P(X=20)\\ &= 0.15+0.10+0.32\\ &= 0.57 \end{aligned}$

d.

$\begin{aligned} \text{Expected value} = E(X) &= \sum_{x} X P(X)\\ & = 18*0.15+19*0.10+20*0.32~+~\\ & ~~~~~~~~~~~~~21*0.05+22*0.13+23*0.25\\ &=20.66 \end{aligned}$

e. $\text{Variance}(X) = E(X^2) - (E(X))^2$ . We caluculate $E(X^2)$ .

$\begin{aligned} E(X^2) &= \sum_{x} X^2 P(X)\\ & = 18^2*0.15+19^2*0.10+20^2*0.32~+~\\ & ~~~~~~~~~~~~~21^2*0.05+22^2*0.13+23^2*0.25\\ &=429.92 \end{aligned}$

$\therefore \text{Variance}(X) = 429.92-20.66^2 = 429.92-426.8356 = 3.0844$

f. $\text{Standard deviation}(X) = \sqrt{\text{Var}(X)} = \sqrt{3.0844} = 1.7562$