1. The time taken to complete a project is described by a normal distribution wi

Question

1. The time taken to complete a project is described by a normal distribution with mean of 80 weeks and a standard deviation of 10 weeks. What is the probability the project takes between 81 and 92 weeks to be completed?

0.3451

0.4247

0.6151

0.8451

2. The time taken to complete a project is described by a normal distribution with mean of 80 weeks and a standard deviation of 10 weeks. What is the probability the project is finished in 70 weeks or less?

0.3413

0.1587

0.8413

0.6587

3. The following is NOT true about continuous random variables

The area under each of the curve represent probabilities

Some may be described by uniform and exponential distributions

The entire area under each curve equals one

They are useful to describe a discrete probability distribution

4. A production process produces items such that 10% are always defective. If two items are randomly selected off the production line, what is the probability that exactly one of these is defective, (either the first is defective and the second is not, or the second is defective, and the first is not, but not both) assuming independence

0.01

0.20

0.18

0.09

5. It has been determined that there is a 65% chance an investor will place her money with JMMB if economic conditions remain the same; a 25% chance of placing her money with JMMB if economic conditions decline and a 50% chance of placing her money with JMMB if economic conditions improve. Analysis has placed the probability of economic conditions remaining the same at 50%; of declining at 20% and of improving at 30%. What is the probability the investor will place her money with JMMB? [One point given to everyone]

0.08125

0.0300

0.11125

0.4050

Explanation / Answer

Ans:

1)

z(81)=(81-80)/10=0.1

z(92)=(92-80)/10=1.2

P(0.1<z<1.2)=P(z<1.2)-P(z<0.1)=0.8849-0.5398=0.3451

2)

z=(70-80)/10=-1

P(z<=-1)=0.1587

3)They are useful to describe a discrete probability distribution is not True

4)Binomial distribution n=2,p=0.1

P(x=1)=2C1*0.1*0.9=0.18

5)P(JMMB)=P(JMMB/same)*P(same)+P(JMMB/decline)*P(decline)+P(JMMB/improve)*P(improve)

=0.65*0.5+0.25*0.2+0.50*0.3

=0.325+0.05+0.15

=0.525

(options are incorrect)

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