# Cheating in Exam

3. a) What is the probability that none of these vehicles requires warranty service? -------------------------------------------------

P(x=0) = 12 C0 (0.10)0 (1-0.10)12-0

-------------------------------------------------

= (1) (1) (0.28243)

-------------------------------------------------

=0.28243

-------------------------------------------------

b) What is the probability that exactly nine of these vehicles require warranty service? -------------------------------------------------

P(x=9) = 12 C9 (0.10)9 (1-0.10)12-9

-------------------------------------------------

= (220) (0.000000001) (0.729)

-------------------------------------------------

=0.00000022 X 0.729

-------------------------------------------------

=0.000000016038

-------------------------------------------------

c) What is the probability that greater than nine of these vehicles require warranty service? -------------------------------------------------

P (x > 9) = p (x= 9), p (x=10), p (x =11), p (x = 12).

-------------------------------------------------

= 12 C9 (0.10)9 (1-0.10)12-9+12 C10 (0.10) 10 (1-0.10)12-10+ 12 C11 (0.10) 11 (1-0.10)12-11+ 12 C12 (0.10)12 (1-0.10)12-12 -------------------------------------------------

=0.000000016038+0.000000005346+0.000000000108+0.000000000001 -------------------------------------------------

=0.000000021493

-------------------------------------------------

-------------------------------------------------

-------------------------------------------------

-------------------------------------------------

-------------------------------------------------

d) What is the probability that three and fewer of these vehicles requires warranty service? -------------------------------------------------

P (x≤ = 3) =p (x =3), P(x= 2), p (x=1), p (x = 0)

-------------------------------------------------

12 C3 (0.10)3 (0.9)9+12 C2 (0.10)2(0.9)10 + 12 C1 (0.10)1 (0.9)11 + 12 C0 (0.10)0 (0.9)12 -------------------------------------------------

=0.08523+0.2301+0.3766+0.282

-------------------------------------------------

= 0.97433

-------------------------------------------------

E) What is the mean and variance for these vehicles?

-------------------------------------------------

The mean= (nπ)

-------------------------------------------------

=12(0.1)

-------------------------------------------------

=1.2

-------------------------------------------------

Variance π (1-π)

-------------------------------------------------

= (12) (0.1) (0.9)

-------------------------------------------------

=1.08

-------------------------------------------------

4 a) what percent of the adults spend more than RM 2500 per year on reading and entertainment? -------------------------------------------------

P (x)> 2500= p (z > (2500 – 1994) =506) =1.12=0.3686 =0.5 – 0.3686 =0.1314 -------------------------------------------------

450 450

-------------------------------------------------

b) What percent of the adults spend more than RM 2500 per year on reading and entertainment? -------------------------------------------------

P (2500 -1994) < z < 3000-1994)

-------------------------------------------------

450450

-------------------------------------------------

=p (1.12 < z < 2.24)

-------------------------------------------------

=0.4875-0.3686

-------------------------------------------------

=0.1189

-------------------------------------------------

-------------------------------------------------

-------------------------------------------------

-------------------------------------------------

-------------------------------------------------

1.12 2.24

-------------------------------------------------

-------------------------------------------------

c) What percent of the adults spend more than RM 3000 per year on reading and entertainment?...

Please join StudyMode to read the full document