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**Unformatted text preview: **AKADEMIA LEONA KOŹMIŃSKIEGO
KOZMINSKI UNIVERSITY MANAGERIAL STATISTICS grading scale ID number :
Name ………………………………………………………….
……………………………………
Surname ……………………………………………………….…..
/ capital letters /
………………………………………………………………………..…. Project: max. 30p. 0 – 59
60– 67
68 – 75
76 – 83
84 – 91
92 – 100 grade: 2
grade: 3
grade: 3,5
grade: 4
grade: 4,5
grade: 5 ……………..……..
task
max.
points Task 1 Task 2 Task 3 Task 4 Task 5 Task 6 Task 7 Task 8 together
∑ Task 1:
The costs of long – distance telephone calls for 10 callers
randomly chosen are as follow: 5, 12, 10, 12, 13, 7, 8, 10, 10, 17 ( PLN )
Construct a box – and – whisker plot for this data set. Write the
interpretation of the first quartile value.
Solution: …………………………………………………………………………………………………. ∶ , , , , , , , , , = two middle values = 10+10
2 = 10 PLN I) : 5, 7, 8, 10, 10 II) : 10, 12, 12, 13, 17 = 1 = 8 = 3 = 12 Interpretation:
1 = 8 1 The value 1 = 8 means that for 25% of callers, the value 8 PLN is the
biggest cost of long distance telephone calls and for 75% of callers, the value 8
PLN is the minimum cost of long distance telephone calls.
or:
1⁄ of population - 25% of long-distance telephone calls cost at most 8 PLN (8
4
PLN or less) and 3⁄4 of population - 75% of long-distance telephone calls cost
at least 8 PLN (8 PLN or more)
Graph: 5 8 10 12 17 Task 2:
A scientist wants to know the middle value of the weekly takings by the small
grocery shops in his town.
Let − means weekly takings in small grocery shops (in thousands $)
weekly takings
till 4
4-8
8 - 12 12 - 18 over 18
number of the shops
8
12
30
13
7
Determine the proper measure for him and give the interpretation. till 4
8
8
4-8
12
20
8 - 12
30
50
12 - 18
13
63
over 18
7
70
SUM
70
----Solution: …………………………………………………………………………………………………. = = 70
2 = 35 (, 12] 2 = + 4
( − − ) = 8 +
(35 − 20) = 30 =8+ 2
∙ 15 = 8 + 2 = 10 ℎ. $
15 Conclusions:
Fifty percent - half of grocery shops has at most 10 th. $ of weekly takings and
the second half of the shops (the next fifty percent) has at least 10 th. $ of
weekly takings. Task 3: Having the data below displayed on the histogram, determine the typical
area of variation if it’s known that the average ≈ , points
5 number of
students 4
3
2
1
2 ̇ 2-4
4-6
6-8
8 - 10
10 - 12 2
5
4
2
1
14 3
5
7
9
11
---- 4 6 8 10 12 examinations scores (points) ̇ ̇ − (̇ − ) (̇ − ) 6
25
28
18
11
88 -3,3
-1,3
0,7
2,7
4,7
----- 10,89
1,69
0,49
7,29
22,09
----- 21,78
8,45
1,96
14,58
22,09
68,86 , Solution: ………………………………………………………………………………………………….
3 =
∑ ̇ ∙ = = = , ≈ , points =√
∑(̇ − ) = √
, = √ , ≈ , points = The typical area of variation: ∈ [ − ; + ] . ∈ [ , − , ; , + , ] points
. ∈ [ , ; , ] points
Conclusions:
A typical student has obtained between 4,1 and 8,5 points.
Typical student has an examination scores between 4,1 and 8,5 points.
Task 4:
Suppose that 5 students were asked their high school GPA
average) and their College GPA, with the answers as follow:
Student
HS GPA
College GPA
A
4.2
3.8
B
3.1
4.2
C
4.9
3.5
D
3.5
5.0
E
3.9
3.7 (grade point Is high school and college GPA related according to this data? Use Spearman
correlation coefficient to answer the question. A
B
C
D
E 4,2
3,1
4,9
3,5
3,9 3,8
4,2
3,5
5,0
3,7 4
1
5
2
3 3
4
1
5
2 = − 1
-3
4
-3
1
0 1
9
16
9
1
36
4 Solution: …………………………………………………………………………………………………. ∑ ∙ = −
= −
= −
= −
= ( − )
( − ) ∙ = − , = −, = −, < ⇒
| | = , ⇒ negative correlation strong correlation Conclusions:
There is a strong negative correlation between HS GPA and College GPA.
The negative correlation suggests that the better HS GPA, the worse College
GPA is or conversely.
or:
We can conclude that the higher HS GPA turns into lower College GPA or
conversely. Task 5:
The newspaper reported the following information about the random
variables: − per capita income and − per capita retail sales (both in
thousands dollars): = , . $ , = , . $ ,
̂ = −, + , , = , . $ , = , .
Suppose you plan to open a retail store in a city where the per capita income is
10 thousand dollars. What does the regression line equation forecast for per
capita retail sales? Estimate the utility of the regression line for making
predictions.
Solution: ………………………………………………………………………………………………….
̂ = −, + , ̂ = ̂(=) = −, + , ∙ = −, + , = , ̂(=) = , . $ ± , . $ 5 = = ( , ) = , % = , ∙ % = %
Conclusions:
The regression line equation forecasts , . $ per capita retail sales in case
of 10 thousand dollars per capita income.
The average mistake of this estimation (the standard error) is , . $ .
Per capita income has the influence on per capita retail sales in 85 %.
Regression line is useful for forecasting . Since tends to 1, the regression line
is good for making predictions.
Task 6:
The table gives the annual income value of T&D Company between 1996 and
2000. $ (millions)
1
1996
45
2
1997
48
3
1998
47
4
1999
48
5
2000
52
Determine the average growth rate
Company .
year /−
/
----/
1,067
/
0,976
/
1,021
/
1,083
of the annual income value of T&D Solution: ………………………………………………………………………………………………….
= − √ / ∙ / ∙ / ∙ / ∙ … ∙ /− = − √ / ∙ / ∙ / ∙ / = 6 = √ , ∙ , ∙ , ∙ , = = √ , = √ , = , ≈ , % = ∙ % = , ∙ % = , %
% = , % > 100 % ⇒ ,7 % growth rate from year to year. , % − % = , %
Conclusions:
The average growth rate (from year to year) of annual income value of T&D
Company between the years 2000 – 1996 was 3,7% .
Second possibility: = − √ = √ = √ , = √ , = , ≈ , % = ∙ % = , ∙ % = , %
Task 7:
According to the American Bankers Association, only one in 10 people are
dissatisfied with their local bank. If 7 people are randomly selected, what is the
probability that the number dissatisfied with their local bank is:
a) exactly two, b) at most two c) between one and three, inclusive ?
Solution: …………………………………………………………………………………………………. − probability of success in one trail = , += = , ⇒ − number of Bernoulli trails = , + , = → 7 people () = () = ( ) ∙ ∙ − − number of dissatisfied clients ) − exactly two clients are dissatisfied
7 − number of success in Bernoulli trails = → dissatisfied clients () = ( ) ∙ (, ) ∙ (, ) = , ≈ , probability from the table ) − at most two clients are dissatisfied
() = ( ≤ ) = ( = ) + ( = ) + ( = )
= () + () + () = = ( ) ∙ (, ) ∙ (, ) + ( ) ∙ (, ) ∙ (, ) + ( ) ∙ (, ) ∙ (, ) = = , + , + , ≈ , probabilities from the table ) − number of dissatisfied clients is between one and three
() = ( ≤ ≤ ) = ( = ) + ( = ) + ( = )
= () + () + () = = ( ) ∙ (, ) ∙ (, ) + ( ) ∙ (, ) ∙ (, ) + ( ) ∙ (, ) ∙ (, ) = = , + , + , ≈ , probabilities from the table Task 8:
The random variable has a normal distribution ∶ (, ).
Find the probability: ( < < ).
Solution: …………………………………………………………………………………………………. − has normal distribution ∶ ( , )
8 =, = We must apply standardization:
= → − ∶ (, ) − the variable which has the standard normal distribution ∶ (, ) ( < < ) =
( −
−
−
<
<
)= = (
= ( −
−
< <
)= < < ) = (, < < , ) = = (, ) − (, ) = = , − , = , values from the table 9 ...

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- Fall '14
- Mariola
- Normal Distribution, Correlation and dependence, Pearson product-moment correlation coefficient, Per capita income, Spearman's rank correlation coefficient, long distance