EF 507
PS-Chapter 7
FALL 2008
1. In a recent survey of college professors, it was found that the average amount of
money spent on entertainment each week was normally distributed with a mean of
$95.25 and a standard deviation of $27.32. What is the probability that the average
spending of a sample of 25 randomly-selected professors will exceed $102.50?
A)
B)
C)
D)
0.0918
0.1064
0.3936
0.4082
ANSWER:
A
2. The time it takes to complete the assembly of an electronic component is normally
distributed with a standard deviation of 4.5 minutes. If we select 20 components at
random, what is the probability that the standard deviation for the time of assembly of
these units is less than 3.0 minutes?
A)
<0.05
B)
<0.005
C)
<0.025
D)
<0.01
ANSWER: C
QUESTIONS 3 THROUGH 6 ARE BASED ON THE FOLLOWING
INFORMATION:
Suppose that 20% of all invoices are for amounts greater than $800. A random
sample of 50 invoices is taken.
3.
What is the mean and standard error of the sample proportion of invoices with
amounts in excess of $800?
A)
B)
C)
D)
Mean = 10, Standard error = 0.4472
Mean = 0.20, Standard error = 0.0566
Mean = 0.20, Standard error = 0.0032
Mean = 10, Standard error = 0.0598
ANSWER: B
4. What is the probability that more than 227% of these invoices are for more than
$800?
A)
B)
C)
D)
0.1368
0.2266
0.3632
0.2734
ANSWER:
C
5. Suppose that 15% of all invoices are for amounts greater than $1000. A random
sample of 60 invoices is taken. What is the probability that at least 25% of these
invoices are for more than $800?
A)
B)
0.1894
0.8106
1
C)
D)
0.3106
0.8894
ANSWER:
A
6. Suppose that 15% of all invoices are for amounts greater than $1,000. A random
sample of 60 invoices is taken. What is the probability that between 17% and 23% of
these invoices are for more than $800?
A)
B)
C)
D)
0.2981
0.2019
0.0962
0.4038
ANSWER:
D
7. The number of students using the ATM on campus daily is normally distributed
with a mean of 237.6 and a standard deviation of 26.3. For a random sample of 55
days, what is the probability that ATM usage averaged more than 230 students per
day?
A)
B)
C)
D)
0.9756
0.9483
0.9838
0.9524
ANSWER:
C
8. A sample of size n is selected at random from an infinite population. As n
increases, which of the following statements is true?
A)
The population standard deviation decreases
B)
The standard error of the sample mean decreases
C)
The population standard deviation increases.
D)
The standard error of the sample mean increases
ANSWER: B
9. Why is the Central Limit Theorem important in statistics?
A)
Because for a large sample size n, it says the population is approximately
normal.
B)
Because for a large sample size n, it says the sampling distribution of the
sample mean is approximately normal, regardless of the shape of the population.
C)
Because for any population, it says the sampling distribution of the sample
mean is approximately normal, regardless of the shape of the population.
D)
Because for any sample size n, it says the sampling distribution of the sample
mean is approximately normal.
ANSWER: B
10. If all possible random samples of size n are taken from a population that is not
normally distributed, and the mean of each sample is determined, what can you say
about the sampling distribution of sample means?
A)
It is positively skewed.
B)
It is negatively skewed.
C)
It is approximately normal provided that n is large enough.
D)
None of the above.
ANSWER: C
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QUESTIONS 11 THROUGH 14 ARE BASED ON THE FOLLOWING
INFORMATION:
Suppose you flip a coin four times. For every head, you receive one point, and for
every tail, you lose one point.
11. Develop the sampling distribution for the mean number of points you receive from
flipping four coins.
ANSWER:
x
-1
-0.5
0
0.5
1
P( x )
0.0625
0.25
0.375
0.25
0.0625
12. What is the standard error for the sampling distribution?
ANSWER:
 X  E ( X ) =  xP( x ) = 0
 X2   x 2 P( x )   X2  0.25.
Hence, the standard error for the sampling distribution of sample means =  X = 0.5
13. What is the probability that the mean number of points you receive on four flips is
0.5?
ANSWER:
P( X  0.5) = 0.25
14. What is the probability that the mean number of points you receive on four flips
is 0?
ANSWER:
P( X  0) = 0.375
QUESTIONS 15 AND 16 ARE BASED ON THE FOLLOWING
INFORMATION:
You have recently joined a country club. Suppose that the number of times you
expect to play golf in a month is represented by a normally distributed random
variable with a mean of 10 and a standard deviation of 2.4.
15.
Over the course of the next year, what is the probability that you average more
than 11 games a month?
ANSWER:
  10,   2.4, n  12
P( X  11) = P(Z > 1.44) = 1 – 0.9251 = 0.0749
16.
Over the course of the next year, the probability is 85% that you average less
than how many games per month?
3
ANSWER:
  10,   2.4, n  12
P ( X  a )  0.85  P ( Z 
a  10
)  0.85 
2.4 / 12
a  10
 1.04  a  10.721
0.6928
17. The number of orders that come into a mail-order sales office each month is
normally distributed with a mean of 298 and a standard deviation of 15.4. For a
particular sample size, the probability is 0.2 that the sample mean exceeds 300. How
big must the sample be?
ANSWER:
P( X  300)  0.20  P( Z 
300  298
15.4 / n
)  0.20  P( Z 
2 n
2 n
)  0.20 
 0.84
15.4
15.4
 n  41.84  42
QUESTIONS 18 THROUGH 21 ARE BASED ON THE FOLLOWING
INFORMATION:
It has been found that 62.1% of all unsolicited third class mail delivered to households
goes unread. Over the course of a month, a household receives 100 pieces of
unsolicited mail.
18.
What is the mean of the sample proportion of pieces of unread mail?
ANSWER:
  = P = 0.621
19.
What is the variance of the proportion?
ANSWER:
 2 = P(1-P) / n = (0.621)(0.379)/100 = 0.00235
20.
What is the standard error of the sample proportion?
ANSWER:
 Pˆ  0.00235 = 0.0485
21.
What is the probability that the sample proportion is greater than 0.60?
ANSWER:
P( P̂ > 0.60) = P(Z >- 0.43) = 0.6664
QUESTIONS 22 AND 23 ARE BASED ON THE FOLLOWING
INFORMATION:
The filling machine at a bottling plant is operating correctly when the variance of the
fill amount is equal to 0.3 ounces. Assume that the fill amounts follow a normal
distribution.
22. What is the probability that for a sample of 30 bottles, the sample variance is
greater than 0.5?
4
ANSWER:
  n  1 s 2  29  0.5  
2

P( s > 0.5)= P 
 = P(  >48.33) < 0.01
2

0.3


2
23. The probability is 0.10 that for a sample of 30 bottles, the sample variance is less
than what number?
ANSWER:
  n  1 s 2 29k 
P( s 2  k )  0.10  P 

  0.10
2
0.3 
 
 P   2  96.67k   0.10
 96.67 k  19.768
 k  0.2045
QUESTIONS 24 AND 25 ARE BASED ON THE FOLLOWING
INFORMATION:
The average checking account balance at a local bank is $1,742 with a standard
deviation of $132. A random sample of 70 accounts is selected.
24. What is the probability that the average balance for these accounts is less than
$1,700?
ANSWER:
  1742 and   132

P( X < 1700) = P  Z 

1700  1742 
 = P(Z < -2.66) = 0.0039
132 / 70 
25. The probability is 0.23 that the average balance for these accounts is greater than
what amount?
ANSWER:

P[ X >k) = 0.23 = P  Z 


K  1742 
K  1742 


  P Z 
15.777 
132 / 70 

K  1742
= 0.74  K = $1,753.67.
15.777
26. The time it takes to complete the assembly of an electronic component is normally
distributed with a standard deviation of 4.5 minutes. If we randomly select 20
components, what is the probability that the standard deviation for the time of
assembly of these units was 50% more than the population standard deviation?
ANSWER:
n =20 and   4.5
  n  1 s 2

 s2

P  s  1.50   P  2  2.25   P 
 19  2.25   P   2  42.75   0.005
2


 

5
27. It has been estimated that 53% of all college students change their major at least
once during the course of their college career. Suppose you are told that the sample
proportion for a random sample was 0.48. Furthermore, you are told that the
probability of getting a sample proportion of this size or smaller is 14%. What must
have the sample size been?
ANSWER:
P = 0.53, P̂ =0/48

P Pˆ  0.48  0.14  P  Z 






0.05 

  PZ 
.
0.2491/ n 
 0.53 0.47  / n  
0.48  0.53
0.05
 1.08  n 17 .
0.2491/ n
QUESTIONS 28 THROUGH 31 ARE BASED ON THE FOLLOWING
INFORMATION:
Given a population with mean  =120 and variance  2 = 81, the central limit applies
when the sample size n  25 . A random sample of size n = 36 is obtained.
28. What are the mean and standard deviation of the sampling distribution for the
sample means?
ANSWER: E ( X )    120,  X   / n  9 / 6  1.50
29. What is the probability that the sample mean is greater than 123?
ANSWER: P( X  123)  P(Z  2)  0.50 – 0.4772 = 0.0228
30.
What is the probability that the sample mean is between 118 and 121?
ANSWER: P(118  X  121)  P(1.33  Z  0.67)  0.4082 + 0.2486 = 0.6568
31.
What is the probability that the sample mean is at most 121.5?
ANSWER: P( X  121.5)  P(Z  1.0) = 0.3413 + 0.50 = 0.8413
QUESTIONS 32 THROUGH 35 ARE BASED ON THE FOLLOWING
INFORMATION:
A random sample of size n =25 is obtained from a normally distributed population
with population mean  =200 and variance  2 =100.
32.
What are the mean and standard deviation of the sampling distribution for the
sample means?
ANSWER: E ( X )    200,  X   / n  10 / 5  2.0
33. What is the probability that the sample mean is greater than 203?
ANSWER: P( X  203)  P(Z  1.5)  0.50 – 0.4332 = 0.0668
34.
What is the value of the sample variance such that 5% of the sample variances
would be less than this value?
6
ANSWER:
2
P(s2  k )  0.05  P[(n  1)s 2 /  2  (n  1)k /  2 ]  0.05  P( 24
 0.24k )  0.05
Hence, 0.24k = 13.848 and k = 57.7. Therefore, if the sample variance is 57.7, then
5% of the sample variances would be less than this value
35. What is the value of the sample variance such that 5% of the sample variances
would be greater than this value?
ANSWER:
2
P(s2  k )  0.05  P[(n  1)s 2 /  2  (n  1)k /  2 ]  0.05  P( 24
 0.24k )  0.05
Hence, 0.24k = 36.415 and k = 151.73. Therefore, if the sample variance is 151.73,
then 5% of the sample variances would be greater than this value
QUESTIONS 36 THROUGH 38ARE BASED ON THE FOLLOWING
INFORMATION:
A random sample of 10 stock market mutual funds was taken. Suppose that rates of
returns on the population of stock market mutual funds follow a normal distribution.
36.
The probability is 0.10 that sample variance is bigger than what percentage of
the population variance?
ANSWER:
or 163.16%
P( 92  14.684)  0.10  14.684 = 9 (Difference)  Difference = 1.6316
37.
Find any pair of numbers, a and b, to complete the following sentence: The
probability 0.95 that the sample variance is between a% and b% of the population
variance.
ANSWER: Since P( 92  2.7)  0.025 and P( 92  19.023)  0.025 , then 2.7 = 9a and
19.023 = 9b.
Therefore, a = 0.30 and b = 2.1137, and the probability is 0.95 that the sample
variance is between 30% and 211.37% of the population variance
38. Suppose that a sample of 20 mutual funds had been taken. Without doing the
calculations, indicate how this would change your answer to question 2.
ANSWER: The interval in question 140 will be smaller
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