Question 1
Suppose a population is normally distributed with a mean of 200 and a standard deviation of 40. The probability that x> 190 is:
0.5987
0.4013
0.7671
0.0987

Answer: 0.5987

Mean=M = 200
Standard deviation =s= 40
x= 190
z=(x-M )/s= -0.25 =(190-200)/40
Cumulative Probability corresponding to z= -0.25 is= 0.4013
Or Probability corresponding to x< 190 is Prob(Z)= 0.4013
Therefore probability corresponding to x> 190 is 1-Prob(Z)= 0.5987 =1-0.4013
0r= 59.87%

Question 2
Another name for the normal distribution is:
exponential distribution
Gaussian distribution
regular distribution
healthy distribution

Answer: Gaussian distribution

Question 3
Values are uniformly distributed between 50 and 80. The height of this distribution is:
0.0333
0.0555
0.1285
0.0437

Answer: 0.0333
0.033333333 =1/(80-50)

Question 4
Suppose x is normally distributed with a mean of 56. Eighty percent of the values are greater than 48. The standard deviation is approximately:
9.52
15.38
26.67
40

Answer: 9.52

Mean=M = 56.00
s=standard deviation= ? (to be determined)
80% of the values are greater than x= 48
Z corresponding to 20% (= 100%- 80.%) is -0.8416
z=(x-M )/s or s=(x-M ) /z*s
or s=(48-56)/-0.8416= 9.51

Question 5
Who of the following is not given some credit for developing or discovering the normal distribution?
Pierre-Simon de Laplace
Abraham de Moivre
Karl Gauss
Karl Pearson

Answer: Karl Pearson

Question 6
Suppose gasoline prices are uniformly distributed across the country ranging from $1.55 to $2.10. If the gasoline price at a particular location is randomly selected, the probability that the price is between $1.70 and $1.80 is:
0.1818
0.0182
0.3333
0.4072

Answer: 0.1818
probability= 0.1818 =(1.8-1.7)/(2.1-1.55)

Question 7
Suppose gasoline prices are uniformly distributed across the country ranging from $1.55 to $2.10. The mean gasoline price for this distribution is:
$1.80
$1.83
$1.85
0.0182

Answer: $1.83

Mean = 1.83 =(1.55+2.1)/2

Question 8
Suppose gasoline prices are uniformly distributed across the country ranging from $1.55 to $2.10. If the gasoline price at a particular location is randomly selected, the probability that the price is between $2.15 and $2.40 is:
0
0.0182
0.3333
0.4545

Answer: 0
Since the range is beyond the distribution

Question 9
Problems from which of the following binomial distributions can be worked by the normal curve because the approximation is good enough?
n = 10, p = .50
n = 12, p = .60
n = 13, p = .70
n = 14, p = .80

Answer: n = 12, p = .60

Question 10
Which of the following distributions represent things that are "measured" as opposed to "counted"?
disjoint distributions
metered distributions
discrete distributions
continuous distributions

Answer: discrete distributions

Question 11
The normal curve is sometimes referred to as the:
curve of inflection
bell-shaped curve
de Moivre's curve
unipeaked curve

Answer: bell-shaped curve

Question 12
An appliance store sells washing machines and dryers among other things. The average sale on washers and dryers is $530 with a standard deviation of $100. Suppose that sales figures on washers and dryers are normally distributed. If a sale is randomly selected, the probability that it is between $650 and $700 is:
0.0705
0.3849
0.1652
0.8403

Answer: 0.0705

Mean=M = 530
Standard deviation =s= 100
x1= 650
x2= 700
z1=(x1-M )/s= 1.2 =(650-530)/100
z2=(x2-M )/s= 1.7 =(700-530)/100
Cumulative Probability corresponding to z1= 1.2 is= 0.8849 0r= 88.49%
Cumulative Probability corresponding to z2= 1.7 is= 0.9554 0r= 95.54%

Therefore probability that the value of x will be between x1= 650 and x2= 700
is = 7.05% =95.54%-88.49% or 0.0705

Question 13
Suppose x is normally distributed with a standard deviation of 7. Seventy-one percent of the values are less than 45. The mean is:
48.85
41.15
43.53
46.47

Answer: 41.15

Mean=M = ? (to be determined)
s=standard deviation= 7

71% of the values are less than x= 45
Z corresponding to 71% is 0.5534
z=(x-M )/s or M = s=x-z*s
or M =45-7*0.5534= 41.13

Question 14
In order to work a binomial distribution problem by the normal curve, what must be true?

n?p> 7
0 <=n
square root of (npq) >= 2.75
n> 5 and .4 <p< .6

Answer: n?p> 7

The Normal distribution can be used to approximate Binomial probabilities when n is large and p is close to 0.5. In answer to the question "How large is large?", or "How close is close?", a rule of thumb is that the approximation should only be used when both np>5 and nq>5.

Question 15
Suppose a population is normally distributed with a mean of 200 and a standard deviation of 40. The probability that x< 150 is:
0.1069
0.2276
0.1056
0.3944

Answer: 0.1056

Mean=M = 200
Standard deviation =s= 40
x= 150
z=(x-M )/s= -1.25 =(150-200)/40
Cumulative Probability corresponding to z= -1.25 is= 0.1056
Or Probability corresponding to x< 150.00 is Prob(Z)= 0.1056

Question 16
A researcher is working a binomial problem using a normal curve approximation.
In the binomial problem, the researcher is trying to determine the probability of 51 <x< 56. In working the problem by the normal curve, the solution will be found at:

50.5 <x< 56.5
50.5 x 56.5
51.5 <x< 55.5
50.5 <x< 55.5

Answer: 50.5 <x< 56.5

Question 17
Suppose 41% of all workers in the telecommunications industry are satisfied with their work. If 63 telecommunications workers are randomly selected what is the probability that fewer than 23 are satisfied with their work?
0.1977
0.2743
0.2358
0.3023

Answer: 0.2358
Probability for sample proportions
p= 41.00%
q=1-p= 59.00%
n=sample size= 63
sp=standard error of proportion=square root of (pq/n)= 6.197% =square root of ( 41.% * 59.% / 63)
psample= 36.51% =23/63
z=(pbar-M )/sp -0.7249 =(0.365079-0.41)/0.06197
Cumulative Probability corresponding to z= -0.7249 is= 0.2343

Question 18
A z score is:
the distance a value is from the mean
the number of standard deviations a value is above or below the mean
the probability that a value has in a normal distribution
the peakedness of the curve

Answer: the number of standard deviations a value is above or below the mean

Question 19
An appliance store sells washing machines and dryers among other things. The average sale on washers and dryers is $530 with a standard deviation of $100. Suppose that sales figures on washers and dryers are normally distributed. If a sale is randomly selected, the probability that it is greater than $600 is:
0.758
0.2943
0.258
0.242

Answer: 0.242

Probability for x
Mean=M = 530
Standard deviation =s= 100
x= 600
z=(x-M )/s= 0.7 =(600-530)/100
Cumulative Probability corresponding to z= 0.7 is= 0.758
Or Probability corresponding to x< 600.00 is Prob(Z)= 0.758
Therefore probability corresponding to x> 600.00 is 1-Prob(Z)= 0.242 =1-0.758
0r= 24.20%

Question 20
Suppose 29% of all commuter cars leaving downtown at 5 P.M. are going somewhere other than home. If 45 commuter cars leaving downtown at 5 P.M. are tracked, what is the probability that more than 11 are going somewhere other than home?
0.195
0.2486
0.695
0.7486

Answer: 0.7486

p= 29.00%
q=1-p= 71.00%
n=sample size= 45
sp=standard error of proportion=square root of (pq/n)= 6.764% =square root of ( 29.% * 71.% / 45)
pbar= 24.44% =11/45
z=(pbar-M )/sp -0.6742 =(0.2444-0.29)/0.06764
Cumulative Probability corresponding to z= -0.6742 is= 0.2501
Or Probability corresponding to pbar< 24.44% is Prob(Z)= 0.2501
Therefore probability corresponding to pbar> 24.44% is 1-Prob(Z)= 0.7499 =1-0.2501
0r= 74.99%

Question 21
Values are uniformly distributed between 50 and 80. The probability of x> 75 is:
0.0333
0.3333
0.1667
0

Answer: 0.1667

Probability= 0.1667 =(80-75)/(80-50)

Question 22
Suppose the average hourly wage of a production line worker in a particular industry is normally distributed with a mean of $9.40. Sixty percent of the workers earn less than $10.25. The standard deviation is:
$1.42
$3.40
$8.50
$2.13

Answer: $3.40

Mean=M = 9.40
s=standard deviation= ? (to be determined)

60% of the values are less than x= 10.25
Z corresponding to 60% is 0.2533
z=(x-M )/s or s=(x-M ) /z*s
or s=(10.25-9.4)/0.2533= 3.4

Question 23
A researcher is working a binomial problem using a normal curve approximation.
In the binomial problem, the researcher is trying to determine the probability of x >=13. In working the problem by the normal curve, the solution will be found at:
X >= 12.5
X >= 13.5
12.5 <=x <= 13.5
x> 13

Answer: X >= 12.5

Question 24
Values are uniformly distributed between 50 and 80. The mean of this distribution is:
0.5
65
70
0.0333

Answer: 65 =(50+80)/2

Question 25
Suppose 57% of all shoppers use credit cards for their purchase in department stores. If 80 such shoppers are randomly selected, what is the probability that more than 49 use a credit card?
0.3106
0.1894
0.2578
0.2946

Answer: 0.2578
(the value that is closest)

p= 57.00%
q=1-p= 43.00%
n=sample size= 80
sp=standard error of proportion=square root of (pq/n)= 5.54% =square root of ( 57.% * 43.% / 80)
pbar= 61.25% =49/80
z=(pbar-M )/sp 0.7671 =(0.6125-0.57)/0.0554
Cumulative Probability corresponding to z= 0.7671 is= 0.7785
Or Probability corresponding to pbar< 61.25% is Prob(Z)= 0.7785
Therefore probability corresponding to pbar> 61.25% is 1-Prob(Z)= 0.2215 =1-0.7785
0r= 22.15%