Tma 2 - Business Statistics_031080377

Seek Chang-Zhin ID:031080377 TMA II

BBM203/05 – 3BST1 Business Statistic

1. Continuous random variable represent measured data than can assume any value within a given range or over an interval or intervals. For example, the heights of all people over 21 years of age, the weight of a food item bought in a supermarket. Discrete random variable represent a list of the possible numerical values or count data that could be made. For example, the number of heads obtained in tossing 4 fair coins. 2. Grade Probability P(X=x) b)

c)

d)

e)

3

A

B

C

D

F

0.25

0.30

0.20

0.15

0.10

Probability that student obtained at least a B grade. P (X≥B) = 1 – (PC)} = 1-{P(X=A) +P(X=B)} = 1- {(0.25+0.30)} = 0.45 Probability that student obtained at least a C grade P (X≥C) = P(X=C) + P(X=D) + P(X=F) = 0.20+0.15+0.10 = 0.45 a). P(X=5) = 25C5 (0.5)5 (0.5)20 = 0.001583 b). P(X=6) = 25C6 (0.5)6 (0.5)19 = 0.005278 c). P(X≤10) = P(Z<10.5-25/6.25) = P(Z<-2.32) = P(Z>2.32) = Q(2.32) = 0.01017 d). P(X≤15) = P(Z<15.5-25/6.25) = P(Z<-1.52) = P(Z>1.52) = Q(1.52) = 0.06426

a)

Seek Chang-Zhin ID:031080377 TMA II

e). E(X) = np = [email protected]=12.5 d). Var(X) = npq = [email protected]@0.5 = 6.25 Standard deviation = √Var(X) = √6.25 = 2.5 4. – there are ‘n’ trials - each trial has two possible outcomes, “success” or “failure”. - the probability of success ‘n’ is the same for each trial. - each trial is independent. 5. a) E(X) = 0(0.48)+1(0.35)+2(0.08)+3(0.05)+4(0.04) = 0+0.35+0.16+0.15+0.16 = 0.82 b) Var(X) = E(X2)-{E(X) 2} E(X2) = 0+0.35+0.32+0.30+0.32 = 1.29 Var(X) = 1.29-(0.82)2 = 0.6176 c) Standard Deviation = √Var(X) = √ 0.6176 = 0.7859 6. Given X~N (150,100) a). P(X<155) = P(Z<155-150/10) = P(Z<0.5) = 1-P(Z>0.5) = 1-0.30854 = 0.69146 b). P(145<X<155) = P(145-150/10158) = P(Z>158-150/10) = P(Z>0.8) = 0.21186 d). P(X>K) = 0.1644 K = 0.976

BBM203/05 – 3BST1 Business Statistic

Seek Chang-Zhin ID:031080377 TMA II

e). P(X>K) = 0.90 K = -1.282 7) a). P(X=0) = 25C0 (0.1)0 (0.9)25 = 0.07179 b). P(X<5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) = 0.07179+0.1994+0.2659+0.2265+0.1384 = 0.90199 c). P(X>2) = 1-P(X≤2) = 1-(0.07179+0.1994+0.2659) = 0.46291

BBM203/05 – 3BST1 Business Statistic