, Test Bank For Probability And Statistics For t1 t1 t1 t1 t1 t1
Engineering And The Sciences 8th Ed by Jay L.
t1 t1 t1 t1 t1 t1 t1 t1 t1
Devore.
t1
Chapter 1 – Overview and Descriptive Statistics
t1 t1 t1 t1 t1 t1
SHORT ANSWER
t1
1. Give one possible sample of size 4 from each of the following populations:
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
a. All daily newspapers published in the United States
t1 t1 t1 t1 t1 t1 t1
b. All companies listed on the New York Stock Exchange
t1 t1 t1 t1 t1 t1 t1 t1
c. All students at your college or university
t1 t1 t1 t1 t1 t1
d. All grade point averages of students at your college or university
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
ANS:
a. Houston Chronicle, Des Moines Register, Chicago Tribune, Washington Post
t1 t1 t1 t1 t1 t1 t1 t1
b. Capital One, Campbell Soup, Merrill Lynch, Pulitzer
t1 t1 t1 t1 t1 t1
c. John Anderson, Emily Black, Bill Carter, Kay Davis
t1 t1 t1 t1 t 1 t1 t1
d. 2.58. 2.96, 3.51, 3.69 t1 t1 t1
PTS: t 1 t 1 1
2. A Southern State University system consists of 23 campuses. An administrator wishes to make an
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
inference about the average distance between the hometowns of students and their campuses. Describe and
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
discuss several different sampling methods that might be employed. Would this be an enumerative or an
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
analytic study? Explain your reasoning.
t1 t1 t1 t1 t1
ANS:
One could take a simple random sample of students from all students in the California State University
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
system and ask each student in the sample to report the distance from their hometown to campus.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
Alternatively, the sample could be generated by taking a stratified random sample by taking a simple
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
random sample from each of the 23 campuses and again asking each student in the sample to report the
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
distance from their hometown to campus.
t1 t1 t1 t1 t1 t1
Certain problems might arise with self reporting of distances, such as recording error or poor recall. This study
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
is enumerative because there exists a finite, identifiable population of objects from which to sample.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
PTS: t 1 t 1 1
3. A Michigan city divides naturally into ten district neighborhoods. How might a real estate appraiser select
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
a sample of single-family homes that could be used as a basis for developing an equation to predict
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
appraised value from characteristics such as age, size, number of bathrooms, and distance to the nearest
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
school, and so on? Is the study enumerative or analytic?
t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1
ANS:
One could generate a simple random sample of all single family homes in the city or a stratified random
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
sample by taking a simple random sample from each of the 10 district neighborhoods. From each of the
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
homes in the sample the necessary variables would be collected. This would be an enumerative study
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
because there exists a finite, identifiable population of objects from which to sample.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
, PTS: t 1 t 1 1
4. An experiment was carried out to study how flow rate through a solenoid valve in an automobile‘s
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
pollution-control system depended on three factors: armature lengths, spring load, and bobbin depth.
t1 t1 t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1 t1
Two different levels (low and high) of each factor were chosen, and a single observation on flow was
t 1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
made for each combination of levels.
t1 t1 t1 t1 t1 t1
a. The resulting data set consisted of how many observations?
t1 t1 t1 t1 t1 t1 t1 t1
b. Is this an enumerative or analytic study? Explain your reasoning.
t1 t1 t1 t1 t1 t1 t1 t1 t1
ANS:
a. Number observations equal 2 2 2=8 t1 t1 t1 t1 t 1 t1 t 1
b. This could be called an analytic study because the data would be collected on an existing
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
process. There is no sampling frame.
t1 t1 t1 t1 t1 t1
PTS: t 1 t 1 1
5. The accompanying data specific gravity values for various wood types used in construction .
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
.41 .41 .42 .42. .42 .42 .42 .43 .44
.54 .55 .58 .62 .66 .66 .67 .68 .75
.31 .35 .36 .36 .37 .38 .40 .40 .40
.45 .46 .46 .47 .48 .48 .48 .51 .54
Construct a stem-and-leaf display using repeated stems and comment on any interesting features of the display.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
ANS:
One method of denoting the pairs of stems having equal values is to denote the stem by L, for ‗low‘ and
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
the second stem by H, for ‗high‘. Using this notation, the stem-and-leaf display would appear as
t1 t1 t1 t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1 t1 t1 t1
follows:
t1
3L 1 stem: tenths t1
3H 56678 leaf: t 1 hundredths
4L 000112222234
5L 144
5H 58
6L 2
6H 6678
7L
7H 5
The stem-and-leaf display on the previous page shows that .45 is a good representative
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 value for the t1 t1
data. In addition, the display is not symmetric and appears to be positively skewed.
t1 t 1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t 1 The spread of the t1 t1 t1
data is .75 - .31 = .44, which is .44/.45 = .978 or about 98% of the typical value of
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 .45. This t 1
constitutes a reasonably large amount of variation in the data. The data value .75 is
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1 a possible outlier.
t1 t1
PTS: t 1 t 1 1
6. Temperature transducers of a certain type are shipped in batches of 50.
t1 A sample of 60 batches was t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1
selected, and the number of transducers in each batch not conforming to design specifications was
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
determined, resulting in the following data:
t1 t1 t1 t1 t1 t1
0 4 t 1 t 1 2 t 1 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1
t 1 1 t 1 t 1 3
2 1 t 1 t 1 2 t 1 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
t 1 4 t 1 t 1 0
5 0 t 1 t 1 2 t 1 1 0 6 4 2 1 6 0 3 3 3 6 1 2 3
t 1 3 t 1 t 1 2
, a. Determine frequencies and relative frequencies for the observed values of x = number of
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
nonconforming transducers in a batch. t1 t1 t1 t1
b. What proportion of batches in the sample has at most four nonconforming transducers? What
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
proportion has fewer than four? What proportion has at least four nonconforming units?
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
ANS:
a.
Number Nonconforming t1 Relative Frequency Frequency t1
0 0.117 7
1 0.200 12
2 0.217 13
3 0.233 14
4 0.100 6
5 0.050 3
6 0.050 3
7 0.017 1
8 0.017 1
1.001
The relative t1 t 1 frequencies don’t add up exactly to 1because they have been rounded
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
b. The number of batches with at most 4 nonconforming items is 7+12+13+14+6=52, which is a proportion
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
of 52/60=.867. The proportion of batches with (strictly) fewer than 4 nonconforming items is
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
46/60=.767.
t1
PTS: t 1 t 1 1
7. The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
t1 for each wafer in a sample size 100, resulting in the following frequencies:
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
Number of t1 t 1 particles Frequency Number of particles t1 t 1 Frequency
0 1 8 12
1 2 9 4
2 3 10 5
3 12 11 3
4 11 12 1
5 15 13 2
6 18 14 1
7 10
a. What proportion of the sampled wafers had at least two particles? At least six particles?
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
b. What proportion of the sampled wafers had between four and nine particles, inclusive? Strictly between
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
four and nine particles?
t1 t1 t1 t1
ANS:
a. From this frequency distribution, the proportion of wafers that contained at least two particles is (100-1-2)/100
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
= t1
.97, or 97%. In a similar fashion, the proportion containing at least 6 particles is (100 – 1-2-3-12-11-
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
15)/100 = 56/100 = .56, or 56%. t1 t1 t1 t1 t1 t1
b. The proportion containing between 4 and 9 particles inclusive is (11+15+18+10+12+4)/100 = 70/100
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
= .70, or 70%. The proportion that contain strictly between 4 and 9 (meaning strictly more than 4
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
and strictly less than 9) is (15+ 18+10+12)/100= 55/100 = .55, or 55%.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
Engineering And The Sciences 8th Ed by Jay L.
t1 t1 t1 t1 t1 t1 t1 t1 t1
Devore.
t1
Chapter 1 – Overview and Descriptive Statistics
t1 t1 t1 t1 t1 t1
SHORT ANSWER
t1
1. Give one possible sample of size 4 from each of the following populations:
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
a. All daily newspapers published in the United States
t1 t1 t1 t1 t1 t1 t1
b. All companies listed on the New York Stock Exchange
t1 t1 t1 t1 t1 t1 t1 t1
c. All students at your college or university
t1 t1 t1 t1 t1 t1
d. All grade point averages of students at your college or university
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
ANS:
a. Houston Chronicle, Des Moines Register, Chicago Tribune, Washington Post
t1 t1 t1 t1 t1 t1 t1 t1
b. Capital One, Campbell Soup, Merrill Lynch, Pulitzer
t1 t1 t1 t1 t1 t1
c. John Anderson, Emily Black, Bill Carter, Kay Davis
t1 t1 t1 t1 t 1 t1 t1
d. 2.58. 2.96, 3.51, 3.69 t1 t1 t1
PTS: t 1 t 1 1
2. A Southern State University system consists of 23 campuses. An administrator wishes to make an
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
inference about the average distance between the hometowns of students and their campuses. Describe and
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
discuss several different sampling methods that might be employed. Would this be an enumerative or an
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
analytic study? Explain your reasoning.
t1 t1 t1 t1 t1
ANS:
One could take a simple random sample of students from all students in the California State University
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
system and ask each student in the sample to report the distance from their hometown to campus.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
Alternatively, the sample could be generated by taking a stratified random sample by taking a simple
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
random sample from each of the 23 campuses and again asking each student in the sample to report the
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
distance from their hometown to campus.
t1 t1 t1 t1 t1 t1
Certain problems might arise with self reporting of distances, such as recording error or poor recall. This study
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
is enumerative because there exists a finite, identifiable population of objects from which to sample.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
PTS: t 1 t 1 1
3. A Michigan city divides naturally into ten district neighborhoods. How might a real estate appraiser select
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
a sample of single-family homes that could be used as a basis for developing an equation to predict
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
appraised value from characteristics such as age, size, number of bathrooms, and distance to the nearest
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
school, and so on? Is the study enumerative or analytic?
t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1
ANS:
One could generate a simple random sample of all single family homes in the city or a stratified random
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
sample by taking a simple random sample from each of the 10 district neighborhoods. From each of the
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
homes in the sample the necessary variables would be collected. This would be an enumerative study
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
because there exists a finite, identifiable population of objects from which to sample.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
, PTS: t 1 t 1 1
4. An experiment was carried out to study how flow rate through a solenoid valve in an automobile‘s
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
pollution-control system depended on three factors: armature lengths, spring load, and bobbin depth.
t1 t1 t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1 t1
Two different levels (low and high) of each factor were chosen, and a single observation on flow was
t 1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
made for each combination of levels.
t1 t1 t1 t1 t1 t1
a. The resulting data set consisted of how many observations?
t1 t1 t1 t1 t1 t1 t1 t1
b. Is this an enumerative or analytic study? Explain your reasoning.
t1 t1 t1 t1 t1 t1 t1 t1 t1
ANS:
a. Number observations equal 2 2 2=8 t1 t1 t1 t1 t 1 t1 t 1
b. This could be called an analytic study because the data would be collected on an existing
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
process. There is no sampling frame.
t1 t1 t1 t1 t1 t1
PTS: t 1 t 1 1
5. The accompanying data specific gravity values for various wood types used in construction .
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
.41 .41 .42 .42. .42 .42 .42 .43 .44
.54 .55 .58 .62 .66 .66 .67 .68 .75
.31 .35 .36 .36 .37 .38 .40 .40 .40
.45 .46 .46 .47 .48 .48 .48 .51 .54
Construct a stem-and-leaf display using repeated stems and comment on any interesting features of the display.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
ANS:
One method of denoting the pairs of stems having equal values is to denote the stem by L, for ‗low‘ and
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
the second stem by H, for ‗high‘. Using this notation, the stem-and-leaf display would appear as
t1 t1 t1 t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1 t1 t1 t1
follows:
t1
3L 1 stem: tenths t1
3H 56678 leaf: t 1 hundredths
4L 000112222234
5L 144
5H 58
6L 2
6H 6678
7L
7H 5
The stem-and-leaf display on the previous page shows that .45 is a good representative
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 value for the t1 t1
data. In addition, the display is not symmetric and appears to be positively skewed.
t1 t 1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t 1 The spread of the t1 t1 t1
data is .75 - .31 = .44, which is .44/.45 = .978 or about 98% of the typical value of
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 .45. This t 1
constitutes a reasonably large amount of variation in the data. The data value .75 is
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1 a possible outlier.
t1 t1
PTS: t 1 t 1 1
6. Temperature transducers of a certain type are shipped in batches of 50.
t1 A sample of 60 batches was t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t 1 t1 t1 t1 t1 t1
selected, and the number of transducers in each batch not conforming to design specifications was
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
determined, resulting in the following data:
t1 t1 t1 t1 t1 t1
0 4 t 1 t 1 2 t 1 1 1 3 4 1 2 3 2 2 8 4 5 1 3 1
t 1 1 t 1 t 1 3
2 1 t 1 t 1 2 t 1 1 3 2 0 5 3 3 1 3 2 4 7 0 2 3
t 1 4 t 1 t 1 0
5 0 t 1 t 1 2 t 1 1 0 6 4 2 1 6 0 3 3 3 6 1 2 3
t 1 3 t 1 t 1 2
, a. Determine frequencies and relative frequencies for the observed values of x = number of
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
nonconforming transducers in a batch. t1 t1 t1 t1
b. What proportion of batches in the sample has at most four nonconforming transducers? What
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
proportion has fewer than four? What proportion has at least four nonconforming units?
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
ANS:
a.
Number Nonconforming t1 Relative Frequency Frequency t1
0 0.117 7
1 0.200 12
2 0.217 13
3 0.233 14
4 0.100 6
5 0.050 3
6 0.050 3
7 0.017 1
8 0.017 1
1.001
The relative t1 t 1 frequencies don’t add up exactly to 1because they have been rounded
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
b. The number of batches with at most 4 nonconforming items is 7+12+13+14+6=52, which is a proportion
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
of 52/60=.867. The proportion of batches with (strictly) fewer than 4 nonconforming items is
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
46/60=.767.
t1
PTS: t 1 t 1 1
7. The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
t1 for each wafer in a sample size 100, resulting in the following frequencies:
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
Number of t1 t 1 particles Frequency Number of particles t1 t 1 Frequency
0 1 8 12
1 2 9 4
2 3 10 5
3 12 11 3
4 11 12 1
5 15 13 2
6 18 14 1
7 10
a. What proportion of the sampled wafers had at least two particles? At least six particles?
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
b. What proportion of the sampled wafers had between four and nine particles, inclusive? Strictly between
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
four and nine particles?
t1 t1 t1 t1
ANS:
a. From this frequency distribution, the proportion of wafers that contained at least two particles is (100-1-2)/100
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
= t1
.97, or 97%. In a similar fashion, the proportion containing at least 6 particles is (100 – 1-2-3-12-11-
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
15)/100 = 56/100 = .56, or 56%. t1 t1 t1 t1 t1 t1
b. The proportion containing between 4 and 9 particles inclusive is (11+15+18+10+12+4)/100 = 70/100
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
= .70, or 70%. The proportion that contain strictly between 4 and 9 (meaning strictly more than 4
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
and strictly less than 9) is (15+ 18+10+12)/100= 55/100 = .55, or 55%.
t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1 t1
