Download Math109Quiz1Fall2011 Soln

North Seattle Community College
Fall Quarter 2011
ELEMENTARY STATISTICS
2618 MATH 109 - Section 05,
Chapter 1 and 2 - Practice Quiz 1
STUDENT NAME: __________________________
10th October 2011
QUIZ SCORE: ______________________________
Question 1
Identify the population and/or the sample in the following. Describe the data set. Also decide
whether the numerical values describe population parameters or sample statistics:
a. A recent survey of a sample of 750 college students reported that the average weekly
income for students is $400.
Population: All college students
Sample: 750 college students surveyed
Data set: Data set comprises of weekly incomes of the 750 college students sampled.
Statistic: Average weekly income of students =$400.
b. The 2182 students who accepted admission offers to Northwestern University in 2009
have an average SAT score of 1442.
Population: 2182 students who accepted admission offers to Northwestern University
in 2009
Data set: Data set comprises of SAT scores of each of the 2182 college students in the
population.
Parameter: Average SAT score of the 2182 students = 1442.
Question 2
If the mean of numbers 28, x, 42, 78 and 104 is 62, then what is the mean of 128, 255,
511, 1023 and x?
Mean of 28, x, 42, 78, 104 = 62
28 + x + 42 + 78 + 104
⇒
= 62
5
⇒ x + 252 = 310
⇒ x = 58
Now,
128+255+511+1023+ x 128+255+511+1023+ 58 1975
=
=
= 395
5
5
5
Question 3
Suppose you are given a data set to analyze. The data consist of 1000 observations on one
variable, the height of the subject being interviewed. Assume half the subjects are male
and half are female. The sample mean height of the males is greater than the sample mean
height the females, while the sample standard deviation of the females is greater than the
sample standard deviation of the males. Both histograms are approximately symmetric
and bell-shaped. Draw two histograms, one for males and one for females, to illustrate
these facts?
x-axis in feet and y-axis is the frequency
Mean ht for men is 6ft > mean of women is 5.5ft (approx.)
Std dev for mean < Stdev for women since the spread of women’s hts is > than that for
men’s
Question 4
John received an 86 on a test whose class mean was 70 with a standard deviation of 5.
Samantha received a 75 on a test whose class mean was 60 with a standard deviation of 4.
Which student had relatively the better test score?
z = ( x – µ )/ σ
John’s score xjohn= 86, John’s class mean is ujohn=70 and σjohn=5
Z-score for John is zjohn= (86 – 70)/5 = 3.2
Samantha’s score xsam= 86, Samantha’s class mean is usam=60 and σsam=4
Z-score for John is zsam= (75 – 60)/4 = 3.75
Samantha had a relatively better score since her score was 3.75 standard deviations above
the class mean. John’s score was only 3.2 standard deviations above his class mean.
Question 5:
The following data represents the ages of all the people that participated in a charity
obstacle race from a small village: (15, 15, 15, 17, 17, 17, 17, 17, 19, 19, 19, 19, 23, 23, 23,
29, 29)
I. Complete the following frequency table:
x
Frequen
cy, f
Relative
frequency
Cumulative
Frequency
15
3
0.18
3
45
-4.59
63.16
17
5
0.29
8
85
-2.59
33.49
19
4
0.24
12
76
-0.59
1.38
23
3
0.18
15
69
3.41
34.92
29
2
N=17
0.12
17
58
9.41
177.16
xf
333
(x-µ)
((x-µ)^2)*f
310.12
The data represents a population since it is the ages of ‘all’ the people that participated
in a charity obstacle race from a small village.
II. Compute the following:
a. Mean, Median and Mode
∑ xf = 333 =19.59
Mean =
N
17
Meadian = (N+1)/2 th value in the data set
= (17+1)/2 th value in the data set
= 9th value in the data set
= 19
Mode = x with the greatest frequency = 17
b. Range, Standard deviation and Variance
Range = 29-15 = 14
(x − µ )2 f 310.12
∑
=
Variance =
=18.24
N
17
Standard Deviation = 4.27
c. Five number summary and IQR
Minimum= 15
Q1=(N+1)/4 th value
= 4.5th value OR average 4th and 5th value
= (17+17)/2
= 17
Q2=9th value =
Q3= (3*(N+1)/4) th value
= 13.5th value OR average 13th and 14th value
= (23+23)/2
= 23
Maximum=29
IQR = 23-17=6
d. Find the age range in which at least 75% of the data lies. Use Chebychev’s
theorem since the distribution is not bell-shaped and symmetric.
(Hint: when k=2, proportion of the data that lies within the 2 standard
deviations is 75%)
Using Chebychev’s theorem we know that 75% of the data lies within 2 standard
deviation of the mean age
=(µ-2σ, µ+2σ)
=(19.59-2*4.27, 19.59+2*4.27)
=(11.04, 28.13)
=(15, 28.13) since the minimum value in the population is 15.