Question

3. Probability+ Central Limit Theorem questions:

a. The return on investment is normally distributed with a mean of 10% and a standard deviation of 5%. What is the probability of losing money?

b. An average male drinks 2 liter of water when active outdoor (with a standard deviation of 0.7). An organization is planning for a full day outdoor for 50 men and will bring 110 liter of water. What is the probability that the organization will run out of water? (8)

c. The lifetime of a certain battery is normally distributed with a mean of 10 hours and standard deviation of 1 hour. There are 4 such batteries in the package.

i. What is the probability that the lifetime of all 4 batteries exceed 11 hours?

ii. What is the probability that the total lifetime of all 4 batteries will exceed 44 hours.

Answer 1

a)

for normal distribution z score =(X-μ)/σ
here mean=       μ=	10
std deviation   =σ=	5.0000

probability of losing money:

probability =

P(X<0)

=

P(Z<-2)=

0.0228

b)

here to run out of water average comsumption should be greater than =110/50=2.2 litre

for normal distribution z score =(X-μ)/σ
here mean=       μ=	2
std deviation   =σ=	0.7000
sample size       =n=	50
std error=σx̅=σ/√n=	0.0990

probability =

P(X>2.2)

=

P(Z>2.02)=

1-P(Z<2.02)=

1-0.9783=

0.0217

c)

i) probability that lifetime of one such battery exceed 11 hours=P(X>11)=P(Z>(11-10)/1)=P(Z>1)=0.1587

hence probability that the lifetime of all 4 batteries exceed 11 hours =(0.1587)⁴ =0.0006

ii)

total life time will exceed 44 hours if average battery life will increase 11 hours

here

here mean=       μ=	10
std deviation   =σ=	1.0000
sample size       =n=	4
std error=σx̅=σ/√n=	0.5000

probability =

P(Xbar>11)

=

P(Z>2)=

1-P(Z<2)=

1-0.9772=

0.0228