Download Practice Exam Questions

Solutions To Course Guide for Test #1: Statistics 1024
_____________________________________________________________________________
Practice Exam Questions
_____________________________________________________________________________
A1. To display the distribution of grades on a statistics test (A, B, C, D, F), it would be correct
to use:
a) a bar graph
b) a bar graph or a pie chart
c) a histogram
d) a stemplot
Solution:
The answer is b) because both pie charts and bar graphs can be used for categorical data.
A2. A study of 5000 university graduates recorded the sex of the graduates, the total university
debt and the average annual salary of the university graduates. Which of the following is true?
a) sex, university debt and annual salary are all categorical variables
b) sex, university debt and annual salary are all quantitative variables
c) sex is a categorical variable and university debt and annual salary are quantitative variables
d) sex is a quantitative variable and university debt and annual salary are categorical variables
Solution:
The answer is c).
A3. The following marks were obtained on a mathematics quiz: 34, 56, 78, 89, 23, 66, 65, 77,
99
To make a stemplot of these data, you would use stems:
a) 10, 20, 30, 40, 50, 60, 70, 80, 90
b) 0, 1, 2, 3, 4, 5, 6, 7, 8, 9
c) 20, 30, 50, 60, 70, 80, 90
d) 2, 3, 5, 6, 7, 8, 9
Solution:
You must include all stems, even if there are no data in them
a) can not be correct because 20 on the left of the line and 3 on the right, would mean 203 and
not 23 as required.
The answer is b).
A4. The following stemplot represents the marks of some statistics students on the first test.
The stems are whole percents in tens and the leaves are percents in ones. What is the outlier?
3
4
5
6
7
8
1
246
345
136
2468
a) 3.1%
b) 31%
c) 4%
d) 40%
Solution:
The answer is b).
A5. An M&M’s chocolate fan is interested in studying the colour distribution of the sugar
coating of the chocolate candies. He opens a bag of M&M’s chocolate candies and classifies
the candies according to the colour of the coating. Which of the following are true?
I) The colour of the sugar coating is a categorical variable
II) A bar graph can be used to display the distribution of the colour variable
III) A pie graph can be used to display the distribution of the colour variable
a) I
b) II
c) III
d) all of I,II, III
e) II and III
Solution:
The answer is d).
A6. Environmental Protection Agency regulations require automakers to give the city and
highway gas mileages for their cars. Here are the highway mileages (miles per gallon, MPG)
for 15 car makes.
Make
MPG Make
MPG Make
MPG
BMW
Buick
Chevrolet
Chrysler
Dodge
23
29
27
27
27
Ford
Honda
Hyundai
Infiniti
Lexus
27
28
27
22
23
Mazda
Mercedes-Benz
Nissan
Rolls-Royce
Toyota
30
22
26
15
28
Which of the following graph(s) is (are) appropriate for displaying the distribution for
the gas mileage variable? Check the most appropriate answer.
(i) bar chart and histogram
(ii) bar chart and boxplot
(iii) stemplot and histogram
(iv) pie chart
Solution:
The answer is (iii).
A7. Harris recently installed a spam filter software, but he still saw spam emails in his inbox.
He made a daily record of the number of spam emails that were delivered to his inbox over the
past 20 days. The following is a frequency histogram for his data. The frequency refers to the
number of days.
Histogram of spam email data
6
frequency
5
4
3
2
1
0
20
40
60
80
100
120
# spam emails
Harris also plotted a stemplot for the data. Which of the following is a correct
stemplot for his data? Check only one answer.
(i) Stemplot A
(ii) Stemplot B
(iii) Stemplot C
A. 2|011355
3|01467
4|12479
5|56
6|
7|
8|0
9|
10|5
B. 2|011355
3|01467
4|12479
5|56
8|0
10|5
Solution:
The answer is (i), Stemplot A.
C. 1|05
2|011355
3|01467
4|12479
5|56
6|
7|
8|0
A8. A consumer recorded the price of a loaf of “DiBaggio” Italian bread at her local
supermarket each week in 2007. The data are presented in a histogram below:
Weekly Price of “DiBaggio”loaf, 2007
12
Frequency
10
8
6
4
2
0
1.76
1.92
2.08
2.24
Price ($)
The distribution displayed above would be described as bimodal. True or false?
Solution:
False. There is only one peak.
A9. Given the following pie graph, about what percent of the people travel by car?
Solution:
Cars
Therefore, 52% travel by car.
52%
A10. A survey was conducted in an undergraduate mathematics course. The students were
asked their:
i) age (in years)
ii) height (in metres)
iii) present major (Math, Chemistry, Physics, Sociology, Psychology, Music, Business, Art,
Other
iv) amount of student loans at the present time (in dollars)
v) present annual salary (in dollars)
vi) plans after graduation (Masters, Work, Undecided)
Which of the above are categorical variables and which are quantitative variables?
Solution:
Age, height, amount of student loans and present annual salary are quantitative variables.
Present major and plans after graduation are categorical variables.
A11. The marks in a statistics course containing 10 students are given below:
5
6
7
8
9
46
2358
14
3
2
Which of the following is true?
a) the distribution is right-skewed
b) the lowest mark is 54 %
c) the second highest mark is 83%
d) all of the above
Solution:
The answer is d).
Practice Exam Questions
B1. Calculate the mean, median and mode of the following marks:
34, 44, 67, 98, 78, 56, 66, 78, 88, 67, 50
Solution:
Mode- occurs most often...Therefore, 67 and 78 (bi-modal)
Median- Write numbers in ascending order and take the middle # which is 67.
34, 44, 50, 56, 66, 67, 67, 78, 78, 88, 98
Mean=
⋯
66
B2. Benjamin has four tests all worth 25% each. Benjamin received three marks on tests: 70,
80, and 90. If Benjamin receives an 80 on the fourth test, which of the following is true about
his average mark?
a) the median mark will change, the mean mark will stay the same
b) the mean mark will change, the median mark will stay the same
c) the mean mark and the median mark will both increase
d) the mean mark will increase, and the median mark will stay the same
e) both the median and the mean remain the same
Solution:
Originally, the numbers are 70, 80, 90
Median= 80 (middle # when written in ascending order)
Mean=
80
After adding another mark of 80... the numbers are 70, 80, 80, 90
Median=
Mean=
The answer is e).
80
80
B3. Samuel had four tests in a mathematics course. He scored 80, 75 and 90 on three of the
tests and his average on all four tests was 78.75. What was the mark on the fourth test?
Solution:
80
75
90
78.75
4
80+75+90+x = 315
x = 315 - 90 - 75 - 80
x=70
Therefore, the mark on the fourth test was 70.
B4. A consumer recorded the price of a loaf of “DiBaggio” Italian bread at her local
supermarket each week in 2007. The data are presented in a histogram below:
Weekly Price of “DiBaggio”loaf, 2007
12
Frequency
10
8
6
4
2
0
1.76
1.92
2.08
2.24
Price ($)
Which of the following is the most likely value of the mean of the prices recorded?
(i) $1.80
(ii) $1.90
(iii) $2.00
(iv) $2.10
(v) $2.20
Solution:
Let X be the price of a loaf. Then the mean of X is
  E( X ) 
 xp( x)
x
 1.68 
1
52
 1.76  521  1.84  527  1.92  10
 2.00  11
 2.08  529
52
52
 2.16  526  2.24  525  2.40  521  2.48  521 
The answer is (iii).
B5. Given the data below, what is the median and the mean?
105.28
52
 2.0246 .
5
6
7
8
9
46
2358
14
3
2
Solution:
The numbers represented by the stemplot are 54, 56, 62, 63, 65, 68, 71, 74, 83, 92
Median=
66.5
Mean=
B6.
⋯
68.8
Which of the following statements is true?
(i) When the distribution is skewed to the left, mean > median > mode.
(ii) When the distribution is skewed to the right, mean < median < mode.
(iii) When the distribution is symmetric and unimodal, mean = median = mode.
(iv) When the distribution is symmetric and bimodal, mean = median = mode.
Solution:
The answer is (iii). Think of the Normal distribution as an example.
B7. Find the mean and median given the following stemplot.
3
4
5
6
7
8
1
246
345
136
2468
Solution:
In ascending order, the numbers are 31, 52, 54, 56, 63, 64, 65, 71, 73, 76, 82, 84, 86, 88
68
Median=
Mean=
⋯
67.5
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Practice Exam Questions
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C1. Which of the following is false?
a) the standard deviation is resistant measure
b) the median is a resistant measure
c) the quartiles are resistant measures
d) the five number summary is a resistant measure
Solution:
a) is false.
C2. The mean of a sample is zero. Which of the following must be true?
(Choose all that apply.)
(i) The sample standard deviation is zero.
(ii) The sample variance must be nonzero.
(iii) The IQR is zero.
(iv) The sample total is zero.
(v) The sample median is zero.
Solution:
Since x 
1 n
 xi , if x  0 then
n i 1
n
x
i 1
i
 0 . The answer is (iv).
C3. If the observations in a data set are all equal, then
(i) the variance of the data set equals 0.
(ii) the mean of the data set equals 0.
(iii) the IQR of the data set equals 0.
(iv) both (i) and (iii).
(v) both (ii) and (iii).
Solution:
IQR= Q3 - Q1 = 0 since all data are the same number
Mean = whatever # all of the data are
Variance = 0
There is no dispersion in the data. The answer is (iv).
C4. Environmental Protection Agency regulations require automakers to give the city and
highway gas mileages for their cars. Here are the highway mileages (miles per gallon, MPG)
for 15 car makes.
Make
MPG Make
MPG Make
MPG
BMW
23
Ford
27
Mazda
30
Buick
29
Honda
28
Mercedes-Benz
22
Chevrolet
27
Hyundai
27
Nissan
26
Chrysler
27
Infiniti
22
Rolls-Royce
15
Dodge
27
Lexus
23
Toyota
28
Sketch a boxplot for the highway mileage data (the observations have been arranged
in ascending order for you). Use the scale provided when you draw the plot. Show
your calculations below the plot.
15, 22, 22, 23, 23, 26, 27, 27, 27, 27, 27, 28, 28, 30, 30
mileage per gallon
13
15
17
19
21
23
25
27
29
31
27
29
31
Solution:
mileage per gallon
13
15
17
19
21
min
23
Q1
25
Q2 Q3
max
min  15 , max  30 ;
The number of observations is n  15 (odd), so the median is the
.
15 1
2
 8 th observation  27
Excluding the median, the upper half of the data is: 27, 27, 27, 28, 28, 30, 30;
Q3  median of the above list  28 .
Excluding the median, the lower half of the data is: 15, 22, 22, 23, 23, 26, 27;
Q1  median of the above list  23 .
IQR  Q3  Q1  28  23  5 ;
Q3  1.5  IQR  35.5 ;
Q1  1.5  IQR  15.5 .
The observation 15 is below 15.5, so it is an outlier.
The second smallest observation is 22, and it is not flagged by the outlier rule.
C5. Harris recently installed a spam filter software, but he still saw spam emails in his inbox.
He made a daily record of the number of spam emails that were delivered to his inbox
over the past 20 days. The following is a frequency histogram for his data. The
frequency refers to the number of days.
Histogram of spam email data
6
frequency
5
4
3
2
1
0
20
40
60
80
100
120
# spam emails
(a)
Which of the following is a correct statement about the distribution of the spam
email data? Check only one answer.
(i) The distribution is roughly symmetric, and the mean is about the same as the
median.
(ii) The distribution is skewed, and the mean is larger than the median.
(iii) The distribution is skewed, and the mean is smaller than the median.
Solution:
(ii) The distribution is skewed to the right, so the mean is larger than the median.
(b) What is the percentage of days over the past 20 days that Harris received fewer than
50 spam emails? Check only one answer.
(i) 16%
(ii) 19%
(iii) 50%
(iv) 80%
Solution:
16
 100%  80% . The answer is (iv).
20
(c) What is the third quartile of the number of spam emails? Use the stemplot you have
chosen in part (a) to answer this question. Check only one answer.
(i) 23
(ii) 25
(iii) 47
(iv) 48
Solution:
The upper half of the data set that is above the median consists of the data points
37, 41, 42, 44, 47, 49, 55, 56, 80, 105.
The third quartile is the median of the above list: Q3 
47  49
 48 . The answer is (iv).
2
(d)
Identify any outliers in the data set using the stemplot you have chosen in part (a). It is
given to you that the IQR is 23.
Solution:
There are no outliers in the lower end (the distribution does not have a long left tail).
Q3  1.5 IQR  48  1.5  23  82.5 .
Since 105  82.5 , the data point 105 is an outlier, the only outlier in the data set.
(e) Which of the following pairs of summary statistics best describe the centre and the
spread of the number of spam emails received daily? Check only one answer and
explain briefly.
(i) mean and standard deviation
(ii) mean and IQR
(iii) median and IQR
(iv) median and variance
Solution:
Both the median and IQR are insensitive to outliers, so in the presence of outliers (105), we
should report these two summary statistics. The answer is (iii).
C6. Calculate the mean, median, range, IQR, and sample standard deviation of the following
data set. Find any outliers, if they exist.
24, 46, 49, 51, 64, 64, 67, 81, 88, 89, 97, 103
Solution:
If we look, we can see that the data is in order. There are 12 data points, so the median is
the average of the 6th and 7th data points. Q1 is in between the 3rd and 4th data points, and Q3
is between the 9th and 10th data points.
Q1  50
median  65.5
Q3  88.5
Range  103  24  79
IQR  88.5  50  38.5
We can use the formulas above to calculate the mean and sample standard deviation:
x 
s 
24  46  97  103
 68.58
12
(24  68.58) 2  (46  68.58) 2  (97  68.58) 2  (103  68.58) 2
 23.69
11
We now check for outliers:
Q3  1.5 IQR  88.5  1.5 (38.5)  146.25
Q1  1.5 IQR  50  1.5 (38.5)   7.75
Since no data point is above 146.25 or below -7.75, there are no outliers in the data set.
C7.
A very hardworking chemist performs a chemical reaction 36 times, but he does not
seem to be very satisfied with the outcome (so he might do more!). He weighs the yield of the
reaction every time and sends the weight data to his statistician friend. This friend uses a
stemplot to display the weights (measured in grams) from the 36 reactions.
15
16
17
18
19
20
21
22
23
24
5
014
123559
11456899
0245667899
00137
2445
(a) Describe the shape of the distribution of the weights of the chemical yields.
Solution:
Unimodal (one peak).
Asymmetric – skewed to the left.
Has a suspected outlier ( 155 g ).
(b) The chemist later realizes that he has made a mistake when entering the data. The
smallest yield of his 36 reactions should be 195 grams instead of 155 grams. Based
on the wrong data, his statistician friend calculates the mean weight to be 218 grams.
(i) Find the correct value of the mean weight.
Solution:
With the wrong data:
With the correct data:
155  190  191   244  245
 218
36
195  190  191   244  245
x 
36
x 

195 155  190  191   244  245 155


36
36
36

195
155
 218 
 219.11 grams
36
36
(ii) How will the values of the following summary statistics change after the data
correction is made?
Choose your answer from the following list:
I. The value becomes smaller after the correction is made.
II. The value becomes larger after the correction is made.
III. There is no change in the value after the correction is made.
Summary Statistics
Your answer is (I, II or III?)
median
______________________
IQR
______________________
standard deviation
______________________
75th percentile
______________________
Solution:
Summary Statistics
median
III
IQR
III
standard deviation
I
th
75 percentile
III
C8. Calculate the mean, median, mode, range, standard deviation and variance of the following
marks: 50, 65, 70, 55, 80, 95, 35, 65, 80
Solution:
Write the numbers in ascending order: n=9
35, 50, 55, 65, 65, 70, 80, 80, 95
Mode=65
Median=middle # = 5th number = 65
Mean=
⋯
66
Range= max - min= 95 - 35 = 60
Make a chart to find the standard deviation:
̅ 2
̅
35-66
961
50-66
256
55-66
121
65-66
1
65-66
1
70-66
16
80-66
196
80-66
196
95-66
841
2589
8
̅
1
17.99
C9. Find the five-number summary and draw a box-plot for the following list of marks:
55, 76, 88, 56, 77, 89, 66
Solution:
55, 56, 66, 76, 77, 88
Min= 55
Max= 88
Q1= 56
Q3=77
Median=
71
C10. If a distribution is skewed to the right, which of the following is true?
a) the mean is equal to the median
b) the mean is less than the median
c) the mean is greater than the median
d) the mean and the mode are equal
Solution:
The answer is c).
C11. What percentage of the observations in a distribution lie between the first quartile and the
third quartile?
a) 25%
b) 50%
c) 75%
d) 100%
Solution:
The answer is b).
C12. What percentage of the observations in a distribution lie before the first quartile?
a) 25%
b) 50%
c) 75%
d) 100%
Solution:
The answer is a).
C13. Which of the following represent possible values of the standard deviation for a set of
data?
a) 0,1.5,2.3
b) -1, 0,1, 2
c) -2, -1, 0, 1, 2
d) -1, 0 , 1.2
Solution:
The standard deviation is always greater or equal to 0.
The answer is a).
C14. Find the interquartile range for the following data and determine if there is a suspected
outlier.
30, 59, 66, 56, 67, 62, 78, 81, 82
Solution:
30, 56, 59, 62, 66, 67, 78, 81, 82
IQR= Q3 - Q1
First, we need the median and then the 1st and 3rd quartiles
Median=66
Q1=middle of 56 and 59=
Q3=
=57.5
79.5
IQR=Q3-Q1=79.5 - 57.5=22
To find outliers
Q3 + 1.5 IQR= 79.5 + 1.5(22) = 112.5
Q1 + 1.5IQR= 57.5 - 1.5(22)= 24.5
An outlier is any number above 112.5 or below 24.5.
Therefore, there are no outliers.
C15. Find the standard deviation and the variance for the following set of data of temperatures
during the summer in London, Ontario:
34, 32, 26, 28, 27, 40, 23, 21, 34, 37, 38, 41
Solution:
21, 23, 26, 27, 28, 32, 34, 34, 37, 38, 40, 41
̅
21
23 ⋯
12
41
31.75
Make a chart to find the standard deviation.
̅
21-31.75
23-31.75
26-31.75
27-31.75
28-31.75
32-31.75
34-31.75
34-31.75
37-31.75
38-31.75
40-31.75
41-31.75
478.16
11
The variance is
̅
115.56
76.56
33.06
22.56
14.06
0.06
5.06
5.06
27.56
39.06
68.06
85.56
6.6
43.5
_____________________________________________________________________________
Practice Exam Questions
_____________________________________________________________________________
D1. For the standard normal random variable Z, find the value of Pr[Z<1.8].
A. 0.0359
B. 0.9641
C. 0.8599
D. 0.1401
E. 0.7881
Solution:
Pr (Z<1.8)=0.9641
The answer is b).
D2. For the standard normal random variable Z, find the value of Pr[Z>-0.60].
A.0.2743
B. 0.7257
C. 0.6
D. 0.5239
E. 0.4761
Solution:
same area as
Pr(Z> - 0.60) = Pr(Z<0.60) = 0.7257
The answer is b).
D3. For the standard normal random variable Z, find the value of
Pr[-1.10<Z<1.10].
A. 0.7286
B. 1.10
C. 0.1357
Solution:
Pr (-1.10<Z<1.10)= Pr(Z<1.1) - Pr(Z<-1.1)
=0.8643 - 0.1357
=0.7286
The answer is a).
D. 0.8643
E. 0.8438
D4. Consider the standard normal random variable Z. Which of the following statements is
false?
A. The mean of Z is   0
B. The standard deviation of Z is
 1
C. For any value of k, Pr[Z>-k]=1 -Pr[Z<K]
D. For any value of k, Pr[Z<-k]=1 - Pr[Z>k]
E. Pr[Z<1.00]=0.8413
Solution:
The answer is c).
D5. For the standard normal random variable Z, what is the value of Pr[Z>1.55]?
A.0.9332
B. 0.0668
C. 0.5596
D. 0.0606
E. 0.9394
Solution:
Pr (Z> 1.55)= 1 - Pr(Z< 1.55)
= 1 - 0.9394
=0.0606
The answer is d).
D6. If Z is the standard normal random variable, find Pr[-0.9 < Z < 0.9].
A. 0.8159
B. 0.1841
C. 0.5359
Solution:
Pr (-0.9<Z<0.9)= Pr(Z<0.9) - Pr(Z<-0.9)
=0.8159 - 0.1841
=0.6318
The answer is d).
D. 0.6318
E. 0.4641
D7. Find the value of k if it is known that Pr[k<Z<2.1]= 0.0489 , where Z is the standard normal
random variable.
A. 1.5
B. 0.0489
C. 0.9511
D. 0.9821
E. -1.5
Solution:
Pr(Z<2.1)= 0.9821
o.9821 - 0.0489 = 0.9332
Look up area 0.9332 and you get k=1.5
The answer is a).
D8. Use the table for the standard normal random variable Z to find
Pr[-0.67<Z<1.81].
A. 0.2155
B. 0.7135
C. 0.2163
Solution:
Pr (-0.67<Z<1.81)=Pr(Z<1.81) - Pr(Z<-0.67)
=0.9649 - 0.2514
= 0.7135
The answer is b).
D. 0.9649
E. 0.7486
D9. Use the table for the standard normal random variable Z to find a value of k for which
Pr[Z<k]= 0.0968.
A. -1.85
B. 1.85
C. -1.86
D. -1.3
E. 1.3
Solution:
Pr(Z<k)= 0.0968
Look up on the table and k is negative....k=-1.3
The answer is d).
_____________________________________________________________________________
Practice Exam Questions
_____________________________________________________________________________
D10. X is a normal random variable with mean
 =100 and standard deviation  =10. Find
Pr[X>90].
A. 0.8413
B. 0.1587
Solution:
90
100
10
1
Pr(X>90)=Pr(Z>-1)= Pr(Z<1)=0.8413
C. 0.8289
D. 0.8531
E. 0.1469
D11. X is a normal random variable with mean
 =10 and standard deviation  =2. Find
Pr[12<X<15].
A. 1.5
B. 0.9938
C. 0.8413
D. 0.9332
E. 0.1525
Solution:
Pr(12<X<15)=Pr
=Pr(1<Z<2.5)
=Pr(Z<2.5) - Pr(Z<1)
=0.9938 - 0.8413
=0.1525
The answer is e).
D12. The amount of cola dispensed by an automated drink dispenser is normally distributed
with mean  =500 ml and standard deviation  =10ml. What is the probability that when you
buy a cola from this machine you will receive less than 480ml?
Solution:
Pr(X<480)=Pr(Z<
)=Pr(Z<-2)=0.0228
D13. X is a normal random variable with mean   25 and standard deviation
the Pr[30 < X < 40].
Solution:
 5 . Find
Pr 1
Pr(30<X<40)=Pr(
3
=Pr(Z<3) - Pr(Z<1)
=0.9987 - 0.8413
=0.1574
D14. The scores on a national achievement test are Normally distributed with mean 500 and
standard deviation 100. What percentage of those who took the test had a score greater
than 630?
Solution:
Let X be the test score. Then X ~ N(  , )
with   500 ,   100 . So
X   630  500 

Pr( X  630)  Pr  Z 



100 

 Pr( Z  1.3)  1  Pr( Z  1.3)  1  0.9032  0.0968 .
D15. The IQ’s of a large population of children are Normally distributed with mean 100.4 and
standard deviation 11.6.
(a) What percentage of the children have IQ’s greater than 125?
Solution:
Let X be the child’s IQ. Then X ~ N(  , )
with   100.4 ,   11.6 . So
X   125  100.4 

Pr( X  125)  Pr  Z 



11.6


 Pr( Z  2.12)  1  Pr( Z  2.12)  1  0.9830  0.0170 .
(b) About 90% of the children have IQ’s greater than what value?
Solution:
Pr( Z  z )  0.90 
z 
x


z   1.28 .
x    z    1.28
So x  100.4  1.28  (11.6)  85.6 .
Thus, 90% of the children have IQ’s greater than 85.6.
D16. The length of human pregnancies from conception to birth varies according to a
distribution that is approximately Normal with mean 266 days and standard deviation 16
days.
(a) What percentage of pregnancies last less than 240 days?
Solution:
Let X be the length of the pregnancy in days. Then X ~ N(  , ) with   266 ,   16 .
X   240  266 

Pr( X  240)  Pr  Z 

  Pr( Z  1.625)  0.0526 .
16



So
(b) What percentage of pregnancies last between 240 and 270 days?
Solution:
270  266 
 240  266
Pr(240  X  270)  Pr 
Z 
  Pr(1.625  Z  0.25)
16
16


 Pr( Z  0.25)  Pr( Z  1.625)  0.5987  0.0521  0.5466 .
(c) How long do the longest 20% of pregnancies last?
Solution:
Pr( Z  z )  0.20 
z 
x


z  0.84 .
x    z    0.84
 266  0.84  (16)  279.44 .
Therefore, the longest 20% of pregnancies last more than 279 days.
(d) 95% of all pregnancies last between x=a and x=b days. Find a and b.
Solution:
Pr(a<X<b)=0.95
Pr(Z< k)= 1-0.025 = 0.975
Look up on the table for Z and k=1.96
266
16
1.96
X=297
AND
1.96
266
16
X=234
Therefore, 95% of all pregnancies last between 234 and 297 days.
D17. A set of final examination grades in an introductory statistics course was found to be
Normally distributed with a mean of 73 and a standard deviation of 8.
(a) What is the probability of getting a 91 or less on the exam?
Solution:
Let X be the final grade. Then X ~ N(  , ) with   73 ,   8 . Then
X   91  73 

Pr( X  91)  Pr  Z 

  Pr( Z  2.25)  0.9878 .
8 


(b) What percentage of students scored between 65 and 89?
Solution:
89  73 
 65  73
Pr(65  X  89)  Pr 
Z 
  Pr(1  Z  2)
8 
 8
 Pr( Z  2)  Pr( Z  1)  0.9772  0.1587  0.8185 .
c) Only 5% of the students taking the test scored higher than what grade?
Solution:
Pr( Z  z )  0.05  z  1.645 .
z 
x


x    z    1.645
So x  73  1.645  (8)  86.16 .
Therefore, 5% of the students scored higher than 86%.
(d) If the professor grades on a curve (i.e. gives A’s to the top 10% of the class regardless
of the score), are you better off with a grade of 81 on this exam or a grade of 68 on a
different exam, where the mean is 62 and the standard deviation is 3?
Solution:
Let X be the score on the first exam and let Y be the score on the second exam. Then
X ~ N(  X , X ) with  X  73 ,  X  8 , and Y ~ N( Y , Y ) with Y  62 ,  Y  3 . So

X   X 81  73 

Pr( X  81)  Pr  Z 

8

X



Y  Y 68  62 

Pr(Y  68)  Pr  Z 

3

Y


 Pr( Z  1)  1  Pr( Z  1)
 Pr( Z  2)  1  Pr( Z  2)
 1  0.8413  0.1587
 1  0.9772  0.0228 .
It would be better to get the grade of 68% on the second exam.
D18. The weight of adult male rhesus monkeys is Normally distributed, with a mean of 15
pounds and a standard deviation of 3 pounds. A rhesus monkey is randomly selected.
(a) Find the probability that the monkey’s weight is less than 13 pounds.
Solution:
Let X be the rhesus monkey’s weight in pounds. Then X ~ N(  , ) with   15 ,   3 .
So
X   13  15 

Pr( X  13)  Pr  Z 

  Pr( Z  0.667)  0.2525 .
3 


(b) Find the probability that the weight is between 13 and 17 pounds.
Solution:
17  15 
 13  15
Pr(13  X  17)  Pr 
Z 
  Pr(0.667  Z  0.667)
3 
 3
 Pr(Z  0.667)  Pr(Z  0.667)  0.7475  0.2525  0.4950 .
(c)
Find the probability that the monkey’s weight is more than 17 pounds.
Solution:
17  15 

Pr( X  17)  Pr  Z 
  Pr( Z  0.667)  1  Pr( Z  0.667)
3 

 1  0.7475  0.2525 .
(d) If 50 rhesus monkeys are randomly selected, about how many would you expect to
weigh less than 12 pounds.
Solution:
12  15 

Pr( X  12)  Pr  Z 
  Pr( Z  1)  0.1587 .
3 

We would expect about 16% of the 50 monkeys to weigh less than 12 pounds, i.e. we would
expect about 8 monkeys to weigh less than 12 pounds.
(e) What is the probability of getting exactly 10 of the 50 monkeys (chosen at random)
that weigh less than 12 pounds?
Solution:
Let Y be the number of monkeys out of 50 that weigh less than 12 pounds. Then
Y ~ Bin(n  50 , p) , where p  Pr( X  12)  0.1587 .
n
 50 
So p( y )  Pr(Y  y )    p y (1  p ) n  y    (0.1587) y (0.8413) 50  y .
 y
y
Therefore, the probability of getting exactly 10 of the 50 monkeys that will weigh less than
12 pounds is
 50 
Pr(Y  10)  p(10)    (0.1587)10 (0.8413) 40  0.1036 .
 10 
D19. The time spent (in days) waiting for a heart transplant for patients with type A+ blood can
be approximated by a Normal distribution with a mean of 127 days and a standard
deviation of 23.5 days.
(a) What is the shortest time spent waiting for a heart transplant that would still place a
patient in the top 30% of waiting times?
Solution:
Let X be the waiting time (in days). Then
X ~ N(  , ) with   127 ,   23.5 .
So Pr( Z  z )  0.30 
z 
x


z  0.52 .
x    z    0.52
 127  0.52  (23.5)  139.2 .
Therefore, 30% of the patients must wait for more than 139 days for a heart transplant.
(b) What is the longest time spent waiting for a heart transplant that would still place a
patient in the bottom 10% of waiting times?
Solution:
Pr( Z  z )  0.10  z   1.28 .
z 
x


x    z    1.28
 127  1.28  (23.5)  96.9 .
Therefore, 10% of the patients have to wait for less than 97 days for a heart transplant.
D20. A statistics test has scores that follow a normal distribution with a mean score of 80 and a
standard deviation of 5.
a) What percent of students scored greater than 90?
b) What percent of students scored below 75?
c) If a students score has a z-score of 0.9772, what is the students score on the statistics test?
Solution:
a) Pr(X>90)= Pr
b) Pr(X<75)=Pr
Pr
2
)=Pr(Z<-1)=0.1587
c) Pr(Z<k)=0.9772
Look up the area 0.9772 on the table and k=2.
2
80
5
10 = X - 80
X=90
1
Pr
2
1
0.9772
0.0228
Therefore, the student scored 90.
D21. The monthly income of pharmacists at a particular pharmacy is normally distributed with
a mean of $3700 and a standard deviation of $400. What percentage of pharmacists earn less
than $3000 per month?
Solution:
3700
400
Pr(X<3000)=Pr(Z<
Pr
1.75
0.0401
D22. The scores on a university finite examination are normally distributed with a mean of 65
and a standard deviation of 10. How high must a student score to place in the top 5% of all
students taking this exam?
Solution:
65
10
Pr(Z<k)=0.95
Look up the area 0.95 on the table and k=1.645
1.645
65
10
X=81.5
Therefore, a student must score 81.5
D23. To tell the exact shape of a Normal distribution, you need:
a) the five number summary
b) the mean, median and mode
c) the mean and the median
d) the mean and the standard deviation
Solution:
The answer is d).
D24. The scores of adults on an IQ test are approximately Normal with a mean of 100 and a
standard deviation of 15. What percent of adults are considered gifted, with an IQ of 130 or
higher? What percent of adults have a score between 110 and 120?
Solution:
100
15
Pr(X>130)=Pr(Z>
Pr(110<X<120)=Pr(
=Pr(Z<1.33) - Pr(Z<0.67)
=0.9082 - 0.7486
=0.1596
Pr
2
0.9772
Pr 0.67
1.33
D25. The scores of adults on an IQ test are approximately Normal with a mean of 100 and a
standard deviation of 15. Helen scores 120 on this test. This means she scores higher than
what percent of all adults?
Solution:
100
15
120
100
15
1.33
Pr(Z<1.33)=0.9082
Therefore, she scores higher than 90.8% of all adults.
D26. Height of women aged 20 to 29 are approximately normal with mean 64 inches and
standard deviation 2.7 inches. Find the z-score for a woman who is 5 feet and 10 inches.
Solution:
Her height is 70 inches.
64
2.7
70 64
2.7
The Z-scores is 2.22
2.22