Question

QUESTION 1 (15 marks) Studd Enterprises sells big-screen televisions. A concern of management is the number of televisions sold each day. A recent study revealed the number of days that a given number of televisions were sold. # of TV units sold #of days Answer the questions below. For each part, show your calculations and/or explain briefly how you arrived at your answer, as appropriate or needed. Required: a. Convert the frequency distribution above into a probability distribution (or relative frequency distribution) showing the proportion of days (rather than the number of days) that the number of televisions sold was 0, 1, 2, 3, 4, and 5 respectively. (3 marks) b. Compute the mean of this general discrete probability distribution. (3 marks) c. Compute the standard deviation of this general discrete probability distribution marks) (5 d. What is the probability that exactly 4 televisions will be sold on any given day? (1 mark) e. What is the probability that 2 or more televisions will be sold on any given day? (1 mark) f. What is the probability that less than 2 televisions will be sold on any given day? (1 mark) g. What is the probability that between 1 and 4 televisions inclusive will be sold on any given day? (1 mark)

Answer 1

a. The frequency distribution into a probability distribution is shown as; x f(x) P(x) 0 2 2/40=0.05 44/40=0.1 10 10/40=0.25 12 12/40=0.3 8 8/40=0.2 4 4/40=0.1 Total 40 Therefore, the probability distribution is. X P(x) 0 0.05 0.1 0.25

b. The mean is. P(x) 0 1 2 3 x.P(x) 0.05 0.1 0.25 0.3 0.2 0.1 1 0.1 0.5 0.9 0.8 0.5 2.8 5 Total E(X) = u = xP(x;)= 2.8 c. The standard deviation is, 0.1 Plx) X.P(x) x x ?P(x) 0 0.05 000 0.1 0.25 0.3 0.2 0.1 1 Total 2.8

O=VEx?P(x)=E(X) = 29.5-(2.8) = 1.28841 d. The probability that exactly 4 televisions will be sold on a given day is, 3 P(x) 0 0.05 0.1 0.25 0.3 0.2 5 0.1 P (X = 4) = 0.2 e. The probability that 2 or more televisions will be sold on a given day is, Plx) 0.05 0.1 0.25 0.3 0.2 0.1

P(X> 2) =P (X = 2)+P (X = 3) +P (X = 4)+P (X = 5) = 0.25 +0.3+0.2 +0.1 = 0.85 f. The probability that less than 2 televisions will be sold on a given day is, x P(x) 0 0.05 1 0.1 0.25 0.3 0.2 uw N 0.1 P (X<2)=P(X=0) + P (X=1) = 0.05 +0.1 = 0.15 g. The probability that between 1 and 4 televisions inclusive will be sold on any given day is,

P(x) WNO 0.05 0.1 0.25 0.3 0.2 0.1 5 P(1 3X 54)=P(X = 1) +P X=2) +P (X = 3)+P (X = 4) = 0.1 +0.25 +0.3+0.2 = 0.85