251solnN1 4/8/08 Open this document in 'Page Layout' view!)
N. Statistical Sampling.
1. Definitions.
2. Distribution of
x
and
p
Text 7.11, 7.14 [7.12, 7.15] (7.12, 7.15)
3. The Central Limit Theorem
Text 7.1, 7.5, 7.67 on CD, 7.68, 7.71 [7.1, 7.5, 7.41, 7.42, 7.45] (7.1, 7.5, 7.40, 7.41, 7.44.) N1, N2.
-----------------------------------------------------------------------------------------------------------------------------------------------------------------.
Problems about the sample proportion
In the outline section L, we learned that if p is a sample proportion, E  p   p , Var p  
pq
. In
n
section M, we learned that the Binomial distribution can be approximated by the Normal distribution, so

pq 
x
that p ~ N  p,
where 0  p   1 , x is the observed number of successes in a sample of n and


n
n


.
0  q  1 p  1
Exercise 7.11 [7.12 in 9th]: Out of a sample of 64 people, 64 are classified as successful. We need to find
a) the sample proportion and b) the standard error of the sample proportion.
x 48
 .75 .
Solution: This problem gives x  48 out of n  64 and that p  .7 . a) p  
n 64
b)  p 
pq

n
.7 .3 
64
.00328125  .0573 Remember that E  p   p  .70 .
Exercise 7.14 [7.15 in 9th]: A pollster will forecast your candidate as the winner of an election if your
candidate receives at least 55% of the vote in a sample. Find the probability of this occurring under the
following circumstances: a) n  100, p  50.1% , b) n  100, p  60% , c) n  100, p  49% . Repeat a) – c)
with n  400 . Discuss.

pq 
This Problem asks for P p  .55  when p has various values and n  100 . Because p ~ N  p,
,

n 

p p p p
z

p
pq
n
a) p  .501
b) p  .60
c) p  .49
.55  .501 




.55  .501 
  Pz  0.98   .5  .3365  .1635
P p  .55   P z  .501 .499    P z 

.00250 



100


.55  .60 




.55  .60 
  Pz  1.02   .5  .3461  .8461
P p  .55   P z  .60 .40    P z 

.00240 



100 

.55  .49 




.55  .49 
  Pz  1.20   .5  .3849  .1151
P p  .55   P z  .49 .51   P z 

.002499 



100 

1
251solnN1 4/8/08 Open this document in 'Page Layout' view!)
d) If n  400 , the Instructor’s Solutions Manual provides the answers below.
Increasing the sample size by a factor of 4 decreases the standard error by a factor of 2.
(a)
P(ps > 0.55) = P (Z > 1.96) = 0.0250
(b)
P(ps > 0.55) = P (Z > – 2.04) = 0.9793
(c)
P(ps > 0.55) = P (Z > 2.40) = 0.0082
Problems about the Sample Mean or
x
Exercise 7.1: If x ~ N 100 ,10  and n  25 , find the probability that x is: a) less than 95, b) between 95
and 97.5, c) above 102.2. d) Show a number x.65 that has 65% above it.
Solution:  x 


10

10 2
x  100
 4  2 , so x ~ N 100 ,2 and z 
.
25
2
n
25
Make diagrams!
95  100 

a) Px  95   P z 
  Pz  2.50   .5  .4938  .0062
2


97 .5  100 
 95  100
z
b) P95  x  97 .5  P
  P 2.50  z  1.25 
2
2


 P2.50  z  0  P1.25  z  0  .4938  .3944  .0994
102 .2  100 

c) Px  102 .2  P z 
  Pz  1.10   .5  .3643  .1357
2


101  100 
 99  100
z
d) (In 9th edition only) P99  x  101  P
  P 0.50  z  0.50 
2
2


 P0.50  z  0  P0  z  0.50   .1915  .1915  .3830
d) in 10th edition e) in 9th edition. We want the point x.65 defined by Px  x.65   .65 . Make a
diagram for z , showing zero in the middle, 100% - 65% = 35% above z .65 (which is below zero),
50% above zero and 50% - 65% = 15% between zero and z .65 . Note that, because 35% is below it,
z.65   z.35 . So we check the table to find the value of z..35 , defined by P0  z  z ..35   .15 . The
closest that we can come on the Normal table is P0  z  0.39   .1517 . So z ..35  0.39 . This
means z.65  0.39 and x.65    z.65  100  0.392  99.22. Note that the bottom of the t table
gives a more accurate value of z..35  0.385 .
f) in 9th edition. Redo the problem with n  16 . Remember that for a continuous distribution, “>”
and "" are essentially the same. The Instructor’s Solutions Manual provides the answers below.
x 


10

10 2
 6.25  2.50 . So x ~ N 100 ,2.5
16
n
16
Px  95   Pz  2.00   .0228
(a)
(b)
(c)
(d)
(e)
P95  x  97.5  P2.00  z  1.00   .1359
Px  102 .2  Pz  0.88   .1894
P99  x  101  P0.40  z  0.40   .3108
x.65    z .65  x  100  0.392.50   99.025
2
251solnN1 4/8/08 Open this document in 'Page Layout' view!)
Exercise 7.5: The diameter of Ping-Pong balls manufactured in a large factory is expected to be
approximately normally distributed with a mean of 1.30 inches and a standard deviation of 0.04 inch.
In the 9th edition the author asks what the probability is that a randomly selected Ping-Pong ball will have a
diameter a) less than 1.28 inches. b) What is the probability that the diameter is between 1.31 and 1.33
inches? c) Between what two values (symmetrically distributed around the mean) will 60% of the balls
fall?
Now the author asks parts d) – h), which are parts a) – d) in the 10th edition. However, it is important to
compare the answers to the questions about probabilities for a sample of 1 as above with the answers in d)
– h) below.
If many random samples of 16 balls are selected d) what will be the values of the population mean and
standard error of the mean? e) What distribution will the sample means follow? f) What proportion of the
sample means (or for an individual sample, what is the probability that the sample mean) will be less than
1.28 inches? g) What proportion of the sample means will be between 1.31 and 1.33 inches? h) Between
what two values (symmetrically distributed around the mean) will 60% of the sample means be?
So now in the original, better version of this problem the author asks you to i) compare the answers of (a)
with (f) and (b) with (g). Discuss. j) Explain the differences between the results in (c) and (h). k) Which is
more likely to occur – an individual ball above 1.34 inches, a sample mean above 1.32 inches in a sample
of size 4, or a sample mean above 1.31 inches in a sample of size 16? Explain.
Note the follow-up below.
Solution: If we have x ~ N 1.30,0.04  and a) – c) are absolutely straightforward Normal distribution
problems.
1.28  1.30 

 Pz  0.50   .5  .1915  .3085
(a) Px  1.28   P  z 
0.04 

(b) P(1.31 < x < 1.33) = P(0.25 < z < 0.75) = 0.2734 – 0.0987 = 0.1747
(c) A symmetrical region around the mean with 60% probability. Make a diagram for z ,
showing zero in the middle. The area we want can be split in two by zero, so that
P0  z  z .20   .30 . If we look for a probability of .30 on the Normal table, the closest we can
come is P0  0.84   .2995 . Our two values of z are z.20  0.84 , and these can be made
values of x by using x    z.20 x  1.30  0.840.04   1.30  .0336 or 1.2664 to 1.3336, and
we can show that P(1.27 < x < 1.34)  .60. If we use the bottom of the t table, z .20  0.842 , a more
accurate value.

0.04 2
 0.0001  0.01 .
16
n
16
(e) Because the population diameter of Ping-Pong balls is approximately normally distributed, the
sampling distribution of samples of 16 will also be approximately normally distributed.
x ~ N 1.30,0.01 .
(f) P( x < 1.28) = P( z < -2.00) = .5 - 0.4772 = 0.0228
(d) If n  16 ,  x    1.30 . So  x 
.

0.04

(g) P(1.31 < x < 1.33) = P(1.00 < z < 3.00) = .4987 – .3413 = 0.1574
(h) A symmetrical region around the mean with 60% probability. We already know that
z .20  0.842 . Our interval will be x    z.20 x  1.30  0.842 0.01  1.30  .0084 or 1.2916
to 1.3084.
(i) When samples of size 16 are taken rather than individual values (samples of n = 1), more
values lie closer to the mean and fewer values lie farther away from the mean with the
increased sample size. This occurs because the standard deviation of the sampling
distribution, the standard error, is given by  x 

. As n increases, the value of the
n
denominator increases, resulting in a smaller value of the overall fraction.
(j) The standard error for the distribution of sample means of size 16 is 1/4 of the population
standard deviation of individual values and means that the sampling distribution is more
concentrated around the population mean.
3
251solnN1 4/8/08 Open this document in 'Page Layout' view!)
(k) They are equally likely to occur (probability = 0.1587) since as n increases, more sample
means will be closer to the mean of the distribution.
Follow up: From what you have learned previously if a random sample of 16 balls is selected, l) what is
the probability that l) all 16 balls will have a diameter less than 1.28 inches? m) What is
the probability that at least one ball will be between 1.31 and 1.33 inches?
Solution: These are binomial problems.
(l) In a) we found Px  1.28   .3085 . If y is the number of balls out of 16 with diameters below
1.28 inches, it has the binomial distribution with n  16 , p  .3085 and q  1  p  .6915. So
16 16 0
P y  16   C16
p q  .3085 16  6.73104 10 9  .0000000067 .
(m) In b) we found P(1.31 < x < 1.33) = 0.1747. If y is the number of balls out of 16 with
diameters between 1.31 and 1.33 inches, it has the binomial distribution with n  16 , p  .1747
and q  1  p  .8253. So P y  1  1  P y  0  1  C 016 p 0 q16  1  .8253 16  1  .0467  .9533 .
This is a great variation on the previous problem for a final exam.
Exercise 7.67 on CD [7.41 in 9th] (7.40 in 8th): Given that N  80 and n  10 and the sample is obtained
without replacement, determine the finite population correction factor.
Solution: If we have N  80 and n  10, the finite population correction factor is
N n
80  10


N 1
80  1
70
 .9413 .
79
Exercise 7.68 on CD [7.42 in 9th] (7.41 in 8th ): Which of the following finite population factors will have
a greater effect in reducing the standard error – one based on a sample of size 100 selected without
replacement from a population of size 400 or one based on a sample of size 400 selected without
replacement from a population of size 900? Explain.
Solution: For N  400 and n  100 ,
N n
900  200


N 1
900  1
part of the population.
n  200 ,
N n

N 1
400  100
300

 .8671 . For N  900 and
400  1
399
700
 .8824 . The first is more effective because the sample is a larger
899
Exercise 7.71 on CD [7.45 in 9th] (7.44 in 8th):
The amount of time a bank teller spends with each customer has a population mean   3.10 minutes and
a standard deviation   0.40 minute. If a random sample of 16 customers is selected without replacement
from a population of 500 customers, a) what is the probability that the average time spent per customer is
less than 3 minutes? b). There is an 85% chance that the sample mean will be below how many minutes?
Solution: x ~ N 3.10,0.40  . N  500 ,
.40 2 484 
N  n 0.40 500  16


 0.0096994  .0985 Note that since the
16 499 
n N 1
16 500  1
sample size is less than 5% of the population, the finite population correction has almost no effect. This
still shows how to use the correction even if it is not needed. x ~ N 3.10,0.0985 
n  16 .  x 
x
3  3.10 

a) Px  3  P z 
  Pz  1.02   Pz  0  P 1.02  z  0  .5  .3438  .1562 .
.0985 

b) We want the 85th percentile, z..15 , a point with 15% above it and 85% below it. Make a diagram and
show that P0  z  z.15   .3500 . If we use the Normal table, the closest we can come is P0  z  1.04 
 .3508 . So z.15 .  1.04. If we use the t table we get z .15  1.036 . Finally x.15    z.15
 3.10  1.036 .0985   3.202 .
4
251solnN1 4/8/08 Open this document in 'Page Layout' view!)
3.202  3.10 

You should be able to show that Px  3.202   P z 
  Pz  1.04 
0.0985 

 .5  .3508  .8508  85%.
Problem N1: The average life of a Toyota Caramba automobile is 44 months with a standard deviation of
18 months.
a) From a sample of 36, what is the probability that we find an average life below 38 months?
Solution: x ~ N 44, 18  , n  36 Remember! The sample mean is Normally distributed if the parent
distribution is Normal, and approximately normally distributed for large n . The mean is E x    , and
X
X
N n
n
n N 1
if the sample is more than 5% of the population. In this part of the problem we can assume that it is

18
18


 3 . So x ~ N 44, 3 and Make a
approximately true that x ~ N  ,  and  x 
n
6
n
36
diagram for this one - you can reuse it for b)
the standard deviation is  x 
if the sample is less than 5% of the population or  x 
 
b) Actually only 200 Toyota Carambas were ever produced. Redo part a) continuing to assume a sample of
36.
Solution: x ~ N 44, 18  , n  36 , but this time N  200 and since n is more than 5% on the population
size, we must use the finite population correction factor. Thus  x 

n
N n
18

N 1
36
200  36
200  1
38  44 

 2.72343 and x ~ N 44, 2.72343  . So Px  38   P z 
  Pz  2.20 
2
.72343 

 Pz  0  P2.20  z  0  .5  .4861  .0139 .
c) An average package weighs 44 lbs. with a standard deviation of 18 lbs. A Toyota Caramba will carry
1368 lbs. If I must deliver 36 packages, what is the chance that my vehicle will not be overloaded? Once
x . We want P
x  1368 . But if
again x ~ N 44, 18  , but this is a birthday party problem about

we take 36 packages, this is the same as saying that the mean is less than
for Px  38  , and the answer is the same as to part a)


1368
 38 . So we are looking
36
Problem N2: For my national fleet (all the same vehicle), mean weekly gasoline consumption is 16.9
gallons with a standard deviation of 3.2 gallons. In a local garage I have 875 gallons of gas and 50 vehicles.
What is the probability of running out of gas this week?
Solution: If we have 50 vehicles, the mean gas consumption is approximately Normally distributed. This is
x is less than 875. But for 50 vehicles
a birthday party problem; we want to know the probability that

875
 17 .5 . For x , the standard deviation
this is the same as saving that the average gas use is less than
50

3.2

 0.4525 . So
(called the standard error of the mean) is  x 
n
50
P
 16 .9 
 x  875   Px  17.5  P z  170.5.4525



 Pz  1.33 
 Pz  0  P0  z  1.33  .5  .4082  .0913
5