Review Questions for Quiz 3
1.
In
a certain city district, the need for money to buy drugs is given as the reason
for 75% of all thefts. Find the
probability that, out of the next 5 theft cases reported in this district
(a)
exactly
2 resulted from the need for money to buy drugs
(b)
at
most 3 resulted from the need for money to buy drugs.
Solution:
This problem uses the binomial distribution.
Success = theft results from the need for money to
buy drugs
Failure = theft results from some other reason
n = total number of trials = 5
R = # of successes out of n
Pr(Success) = p = 0.75
b(r; n, p) = n!/{r!(n-r)!} pr (1- p)n-r
(a)
Pr(R=2)
= b(2; 5, 0.75) =
5!/{2!(5-2)!}* 0.752 *(1-0.75)5-2 = 10*0.5625*0.0156 = 0.0878
(b) Pr(R<=3) = Pr(R=1)+Pr(R=2)+Pr(R=3) =
b(1; 5, 0.75) + b(2; 5,0.75) + b(3; 5, 0.75) =
0.0146 + 0.0878 + 0.2637 = 0.3661
2.
For
each of the following situations, explain why you can OR cannot use the normal
distribution as the sampling distribution for the sample mean.
a. As reported by Runner’s World magazine, the times of finishers in the New York City 10-K run are normally distributed with a mean of 61 minutes and a standard deviation of 9 minutes. You take a sample of 4 finishers and want to know the probability that their mean time is greater than 65 minutes.
b. According to the U.S. Census Bureau, the mean annual alimony income received by women is $8657 with a standard deviation of $7500. The distribution of alimony income for women is not known. You take a sample of 100 women and want to know the probability that their average alimony income is less than $8000.
c. A brand of water-softener comes in packages marked “net weight 40 lb.” The company that packages the water-softener claims that the bags contain an average of 40 pounds of product and that the standard deviation of the weights is 1.5 pounds. The distribution of the weights is not known. You take a sample of 3 bags of water-softener and want to know the probability that the average weight is less than 40 pounds.
Solution:
(a) Yes. The population is normally distributed. Therefore, the sample mean will be normally distributed.
(b) Yes. Although the population distribution is not known, the sample size is 100. Therefore, since we have a large sample size, the Central Limit Theorem tells us that the sample mean will be approximately normally distributed.
(c) No. The population distribution is not known AND the sample size is very small (n=3). Therefore, neither of our rules applies, so we cannot say that the sample mean is normally distributed.
3. Actually calculate the probability requested in question 3a). That is, what is the probability that the average finishing time for the race is greater than 65 minutes?
Solution:
We have already established that the sample mean will be normally distributed. Specifically, it will have a normal distribution with
Mean = 61 minutes (population mean)
Std. Dev. = 9 / 2 = 4.5 (pop. std. dev. divided by the square root of n)
So:
Pr(x-bar>=65) = Pr{(x-bar-61)/4.5 >= (65-61)/4.5} = Pr(Z>=0.89)
= 1 – Pr(Z<=0.89) = 1-0.8133 = 0.1867
4.
For
the following questions, assume that you have a r.v., Z, that has a standard
normal distribution. Sketch a graph and
give a numerical answer for each question.
a.
What
is the probability that Z is less than 1.63?
b.
What
is the value z such that 87.08% of the observations are less than z (e.g., find
the 87.08th percentile)?
Solution:
a.
Pr(Z<=1.63) = 0.9484
b. Look up 0.8708 in the body of Table D. You will find z=1.13.
5. A lawyer commutes daily from his suburban home
to his midtown office. Assume that the distribution of his trip times is
normally distributed with a mean of 24 minutes and a standard deviation of 3.8
minutes. What is the probability that one of his trips will take more than 30
minutes?
Solution:
Let X = length of
one trip to the office. X~N(24, 3.8)
Pr(X>=30)
= Pr{(X-24)/3.8 >= (30-24)/3.8)} = Pr(Z>= 1.58)
= 1 – Pr(Z<=1.58) = 1 –
0.9429 = 0.0571
