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Economics 202
Spring 2007
Homework 6
Your homework is due at the beginning of class Wednesday, 3/7/07. Homework must be
legible, stapled, and must have your name on it. No late homework will be accepted
under any circumstances.
Book problems:
5.54 (a, b, and c only), 5.102, 5.104
5.54
a. z = (11.3-18)/4 = 1.68
P(x>11.3) = P(z >1.68) = .5 + .4535 = 0.9535
b. z = (2-18)/4
P(x<2) = P(z<-4.0) = .5 - .5 = 0
c. z = (22-19)/4 = 1
P(x>22) = P(z>1) = .5-.3413 = .1587
The chance of waiting more than 22 minutes is about 16%. This is not high, but certainly it is not
extremely unusual that a customer would have to wait this long.
d. If they wanted the wait time to be no more than 18 minutes, the standard deviation would have to be
zero, which is essentially impossible. Thus they would have to try to reduce the mean wait time to the
point where the chance of waiting 18 minutes was virtually zero... this would be a mean of about 4 minutes.
5.102
a. z = (18-18)/6 = 0
P(x>18) = P(z>0) = .5
b. z1 = (10-18)/6 = -1.33
z2 = (20-18)/6 = 0.33
P(10<x<20) = P(-1.33<z<.33) = 0.1293+0.4082 = 0.5375
c. z = -1.33 (from above)
P(x<10) = P(z<-1.33) = .5 - .4082 = 0.0918
d. There’s probably more than one correct way to think about this, but one way to do it is to think about
what average usage by the population of 14,500 could be before you run out of water. That would be
350,000/14,500 = 24.1379. Then figure out the probability that usage is greater than 24, which is 0.1587.
Thus there is approximately a 16% chance of running out of water. Another way to do it is to multiply
average usage (18) by the population, which gives the average total usage. Then figure out where 350,000
is relative to total usage, which again gives you about a 16% chance of running out of water.
5.104
a. z = (2-1.5)/0.25 = 2
P(x>2) = P(z>2) = .5-.4772 = 0.0228
b. z1 = (1-1.5)/0.25 = -2
z2 = (2-1.5)/0.25 = 2
P(1<x<2) = P(-2<z<2) = .4772 + .4772 = .9544
c. z = (1.5-1.5)/.25 = 0
P(x<1.5) = P(z<0) = .5 for one steer
Assuming independence, probability for two steers = (.5)(.5) = 0.25
1. A population is known to be normally distributed with a mean of 100 and standard deviation of 50.
a. What is the probability that a randomly selected item from this population will have a value of 50 or
greater?
Z = (50-100)/50 = -1
P(z>-1) = .3413 + .5 = .8413
b. If a random sample of 25 is taken, what is the probability that the sample mean will be 50 or greater?
Z = (50-100)/(50/5) = -5
P(z>-5) = 1
c. If a random sample of 100 is taken, what is the probability that the sample mean will be 50 or greater?
Z = (50-100)/(50/10) = -10
P(z>-5) = 1
d. Carefully draw a graph showing the population distribution, the sampling distribution of the mean if the
sample size is 50, and the sampling distribution of the mean if the sample size is 100. Draw your graph as
carefully as possible, and then explain, with reference to the graph, why your answers to a, b, and c are
different.
Graph (not shown here) should show three NORMAL distributions of different width and height. The
population should have the greatest width and least height; the sample of 100 should have the least width
and the greatest height.
The probabilities on b. and c. are not different, but the idea is that as the sample size increases, the
standard error (standard deviation of the sample mean) decreases, and thus the distribution is less
variable. Therefore, the probability of getting an extreme result is lower (or the probability of getting a
result close to the mean is higher).