Problem Solving 11/01/2007

1. T/F There is a 50% chance of rain today and a 50% chance of rain tomorrow. Therefore, there is a 100% chance of rain either today or tomorrow.

False. The two events are not disjoint to each other, so addition rule can not be applied.

2. T/F If you toss a coin seven times, it is more likely that heads will come up six times and tails once than that heads will come up seven times in a row.

True. Pr[6 heads and 1 tail]==7*.57

> Pr[7 heads and 0 tail]==1*.57

3. T/F Number of correct guesses at 10 true-false questions follows normal distribution, when you randomly guess all answers.

False. We have n=10 and , such that . Thus, the normal approximation can not applied for this case.

4. If 10% of the nails produced by a machine are defective, the probability that out of four nails chosen at random Let X denote the number of defective nails in the sample.

T/F (a)1 will be defective is 0.2916, True. Pr(X=1)==.2916

T/F (b)0 will be defective is 0.6561, True. Pr(X=0)==.6561

T/F (c)at most 2 nails will be defective is 0.9477.

False. Pr(X<=2) =Pr(X=0)+Pr(X=1)+Pr(X=2)>Pr(X=0)+Pr(X=1)=.9477

5. The mean weight of 600 male employees at a certain company is 151 lb and the standard deviation is 15 lb. Assuming that the weights are normally distributed, the expected number of employees that weigh

T/F (a) between 100 and 150 lb are 360

False. Let X denote the weight of randomly selected employee. Since 100 and 150 are both smaller than the mean 151, the proportion between them should be smaller than 50%, and that expected number of employees should be less than 300.

T/F (b) more than 185 lb are 7.

True. , Pr(Zzs)=.0117, and 600* Pr(Zzs) roughly equals 7.

6. T/F The expectation of a discrete random variable Z whose probability function is given by Pr(Z=z) = (1/2)z (for z= 1, 2, 3, …) is E(Z) = 1.

False. Since “1” is the smallest element in the sampling space, and the probabilities for larger values in the space are not zero, E(Z) must be larger than 1.

7. You are presented an opportunity to invest in a wildcat oil drilling operation. The investment requires a $10,000 investment upfront, non-refundable. There is a 50% chance no oil will be found and your payout will be zero. There is a 30% chance that the well will be only yield a small production and your payout will be $5000. There is a 20% chance that the well will be a gusher and then your payout will be $100,000. What are your expected payout and its standard deviation?

Expected payout: $0*50%+$5000*30%+$100000*20%=$21500

SD: =39309.67

Remark:

Ifand, and X is INDEPENDENT from Y, then

and
Example: If, , and X is independent from Y, then:

and.

8. Assume that the heights of 3500 female students at a university are normally distributed with mean 67.0 inches and standard deviation of 3.0 inches. If 80 samples consisting of 25 students are obtained, what would be the expected mean and standard deviation of the resulting sampling distribution of means if sampling were done with replacement?

T/F 67.0 and 0.6 respectively

True. Mean=67inches. Sd==.6inches

9. The electric bulbs of brand A have a mean lifetime of 1500 hours with a standard deviation of 200 hours, while those of brand B have a mean lifetime of 1300 hours with a standard deviation of 100 hours. If random samples of 125 bulbs of each brand are tested, what is the probability that the brand A bulb sample will have a mean lifetime which is at least (a) 160 hours, (b) 250 hours more than sample mean for the brand B bulb sample?

T/F (a)0.7972 True., Pr[Z>-2]=.7972

T/F (b)0.0022 True., Pr[Z>2.5]=.0022

10. The variance of the sum obtained in tossing a pair of fair dice is:

The variance of the average number obtained in tossing a pair of fair dice is:35/24

(a) (b) (c)(d)