PREPARATION FOR FINAL EXAM (STATISTICS FOR BUSINESS)
1. The Los Angeles Times regularly reports the air quality index for variousareas of Southern California. A sample
of air quality index values for Pomona provided the following data: 28.5, 42.35, 58, 48, 45, 55. Use Excel to:
a) Compute the range.
b) Compute the interquartile range.
c) Compute the sample variance.
d) Compute the sample standard deviation.
e) Compute the coefficient of variation.
ANSWER (Question 1)
Measure
Excel explanation (formula)
Range
=max-min=F12-F13
Interquartile range
=Q3-Q1=F12-F13
Sample variance
=VAR(C7:C12)
Sample standard deviation
=STDEV(C7:C12)
Coefficient of variation
=stdev/mean (%)=G10/F16
Result
29.5
10.2375
109.8904
10.48286
22.72%
2. Consider a very large number of students taking a university entrance exam. Suppose the mean score on the
exam is 300 with a standard deviation of 40.
a)Use Excel to find the z-score for a student who scored 376.
Excel explanation and result: =STANDARDIZE(D25,$H$24,$H$25) 1.9
Interpretation: This value is 1.9 standard deviations above than mean.
b)A student is told that his z-score on this test is -2.3. What was his actual exam score?
Answer:
3. Consider a sample with a mean of 30 and a standard deviation of 5. The distribution is not known. Use Chebyshev’s
theorem in Excel to determine and interpret the percentage of the data within each of the following ranges:
a) 15 to 45
Excel explanation and result: =1-(1/E31^2) 88.9%
Interpretation: At least 88.9% of the data is in the range between 15 and 45.
b) 12 to 48
Excel explanation and result: =1-(1/E33^2) 92.3%
Interpretation: At least 92.3% of the data is in the range between 12 and 48.
4. Consider a sample with data values of 10, 20, 12, 17, and 16. Use Excel to compute and interpret the z-score for
first two observations (10 and 20).
Excel explanation and result (10): =STANDARDIZE(B42,$F$41,$F$42) -1.25
Interpretation (10): This value is 1.25 standard deviations below the mean.
Excel explanation and result (20): =STANDARDIZE(B43,$F$41,$F$42) 1.25
Interpretation (20): This value is 1.25 standard deviations above the mean.
1
5. A bowler’s scores for six games were 128, 135, 177, 181, 115, and 136. Use Excel to comment the distribution shape
(skewness).
Excel explanation and result: =SKEW(B60:B65) 0.63
Interpretation: This is right or positive distribution.
6. Liquid detergent cartons are filled automatically on a production line. Filling weights frequently are known to
be bell-shaped. The mean filling weight is 13.5 dl and the standard deviation is 0.38 dl. Draw the conclusions using
empirical rule.
Answer:
7. Consider the following data.
4
6
11
7
13
8
11
13
9
10
12
13
15
6
6
10
14
14
14
16
15
5
9
12
9
10
8
12
9
14
11
12
Develop a frequency distribution using classes of 4–6, 7–9… and compute the sample mean,sample variance and
sample standard deviation for grouped data.
Answer and the corresponding table:
class
freq f
M
Maverage
f*M
(Mf*(Mav/mean)sq av/mean)sq
4-6
5
5
25
-5.5
30.25
151.25
7-9
7
8
56
-2.5
6.25
43.75
10-12
10
11
110
0.5
0.25
2.5
13-16
10
14.5
145
4
16
160
total
32
336
357.5
mean
10.5
variance 11.53226
st dev
3.395918
2
8.The president of Floor Coverings Unlimited wants information concerning the relationship between number of
commercials and sales volume (in hundreds of dollars). He obtained the following random sample on number of
commercials and sales volume: (7,75) (5,61) (4,58) (5,45) (3,21)
The first number for each observation is number of commercials (x), and the second number is sales volume (y).
a) Develop a scatter diagram with x on the horizontal axis in Excel.
Excel explanation: Select x and y, insert scatter, add trendline
Scatter diagram:
sales volume
scatter diagram
90
80
70
60
50
40
30
20
10
0
y
Linear (y)
0
2
4
6
8
number of commercials
b) Calculate and interpret the covariance.
Excel explanation and result: =COVAR(C49:C53,D49:D53) 20.4
Interpretation: There is a positive or direct linear relationship.
c) Calculate and interpret the correlation coefficient.
Excel explanation and result: =CORREL(C49:C53,D49:D53) 0.84
Interpretation: There is a strong positive relationship between x and y.
9.Suppose that we have a sample space 𝑆 = {𝐸1 , 𝐸2 , 𝐸3 , 𝐸4 , 𝐸5 , 𝐸6 } where 𝐸1 , 𝐸2 , … , 𝐸6 denote the sample points.
The following probability assignments apply:
𝑃 (𝐸1 ) = 0.10, 𝑃 (𝐸2 ) = 0.17, 𝑃 (𝐸3 ) = 0.22, 𝑃(𝐸4 ) = 0.31, 𝑃(𝐸5 ) = 0.14, 𝑃 (𝐸6 ) = 0.06.
Let:
𝐴 = {𝐸1 , 𝐸3 , 𝐸4 , 𝐸6 }
𝐵 = {𝐸2 , 𝐸4 , 𝐸5 }
𝐶 = {𝐸2 , 𝐸3 , 𝐸5 }
a) Find 𝑃(𝐴), 𝑃(𝐵) and 𝑃(𝐶).
b) Find 𝐴 ∪ 𝐶 and 𝑃(𝐴 ∪ 𝐶).
c) Find 𝐴 ∩ 𝐶 and 𝑃(𝐴 ∩ 𝐶 ).
d) Are events A and B mutually exclusive.
e) Find 𝐵̅ and 𝑃 (𝐵̅).
Answer:
a) P(A)= 0.1+0.22+0.31+0.06=0.69
P(B)=0.17+0.31+0.14=0.62
P(C)=0.17+0.22+0.14=0.53
b) A U C = (E1, E2, E3, E4, E5, E6)
P(A U C)=P(A)+P(C)-P(A inter C)=0.69+0.53-0.22=1
3
c) A intersection C =(E3)
P (A int C)= P(E3)=0.22
d) A int B is empty scope in that case events are mutually exclusive
A int B = E4 these events are not mutually exclusive
e) B compl = (E1, E3, E6)
P (B compl)= 1-P(B)= 1-0.62=0.38
10.Suppose that we have a sample space with five equally likely experimental outcomes:𝐸1 , 𝐸2 , 𝐸3 , 𝐸4 , 𝐸5 . Let:
𝐴 = {𝐸1 , 𝐸3 }
𝐵 = {𝐸2 , 𝐸4 , 𝐸5 }
𝐶 = {𝐸2 , 𝐸3 , 𝐸5 }
a) Find 𝑃(𝐴), 𝑃(𝐵) and 𝑃(𝐶).
b) Find 𝐵 ∪ 𝐶 and 𝑃(𝐵 ∪ 𝐶).
c) Find 𝐴 ∩ 𝐶 and 𝑃(𝐴 ∩ 𝐶 ).
d) Are events A and B mutually exclusive.
e) Find 𝐶̅ and 𝑃 (𝐶̅ ).
f) Find 𝐴̅ ∪ 𝐵 and 𝑃(𝐴̅ ∪ 𝐵).
Answer:
P(E1)=P(E2)=P(E3)=P(E4)=P(E5)=1/5=0.2
A) P(A)=0.2+0.2=0.4
P(B)=0.2+0.2+0.2=0.6
P(C)=0.2+0.2+0.2=0.6
B) B U C = (E2, E3, E4, E5) P(B U C)= P(B)+P(C)-P(B INT C)=0.6+0.6-0.4=0.8
B INT C = (E2, E5)= 0.2+0.2=0.4
C) A INT C = (E3) P(A INT C)= P(E3)=0.2
D)A INT B = EMPTY SCOPE SO THESE EVENTS ARE MUTUALLY EXCLUSIVE
E)C COMPL = (E1, E4)
P(C COMP)=1-P(C)=1-0.6=0.4
F)A COMP = (E2, E4, E5)
A COM U B = (E2, E4, E5)
P(A COM U B)= P(A COM)+P(B)-P(A COM INT B)=0.6+0.6-0.6=0.6
A COM INT B=(E2, E4, E5)
P(A COM INT B)=0.2+0.2+0.2=0.6
P(A COM)=1-P(A)=1-0.4=0.6
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11. A Paris club obtains the following data on the age and marital status of 140 customers.
a)
b)
c)
d)
e)
Marital status
Age
Single
Married
Under 30
22
8
30 or over
56
33
Develop a joint probability table for these data.
Use the marginal probabilities to comment on the age of customers attending the club.
Use the marginal probabilities to comment on the marital status of customers attending the club.
What is the probability of finding a customer who is married and under the age of 30?
If a customer is 30 or over, what is the probability that he or she is married?
Answer:
a)
Marital status
Single
Married
Age
Under
30
total
0.184874
0.067227 0.252101
30 or
over
total
0.470588
0.655462
0.277311 0.747899
0.344538
1
b)Most of the customers are 30 or more years old (74.8%).
c)Most of the customers are single (65.5%).
d) 0.067=6.7%
e)
STRUCTURE OF FINAL EXAM (IM and IFB)
1. Q1 from preparation (12 p)
2. Q3 from preparation (12 p)
3.Q10 from preparation (16 p)
4.from preparation excluding Q6 (5 p)
5.problem question (15 p)
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