Probability: Mean, Standard Deviation,
Range, and Confidence Interval
Question 1
a) Average (µ) = 42.3 and standard deviation () = 8.9
Proportion of students older than 50 are
P(X ˃ 50) = P((X -
µ)/ ˃(50-42.3)/8.9)
Since Z = (x - µ)/
And (50-42.3)/8.9
= 0.87
Then P (X ˃ 50) = P(Z ˃ 0.87)
P (Z ˃ 0.87) =
0.1922
% of students older than 50 = 19.22%
Out of 200 students in the Excel table, there are 50 who are
over 50 years old. This gives an actual percent of
50/200 X 100 = 25%
This is more than the computed estimate of 19.22%.
b) Average (µ) = 42.3 and standard deviation () = 8.9
Proportion of students younger than 40 are
P(X ˂ 40) = P((X -
µ)/ ˂ (40-42.3)/8.9)
Since Z = (x - µ)/
And (40-42.3)/8.9
= -0.26
Then P (X ˂ 40) = P (Z ˂ -0.26)
P (Z ˂ -0.26) = 0.3974
% of students younger than 40 = 39.74%
Out of the 200 students in the Excel table, there are 89 who
are under 40 years old, giving a value of
89/200 X 100 = 44.5%
This is more than the computed probability of 39.74%.
c) Average (µ) = 42.3 and standard deviation () = 8.9
Proportion of students between 41 and 46 (inclusive) are
P(40 ˂ X ˂ 47) =
P(40-42.3 ˂ X - µ ˂ 47-42.3)
P((40-42.3)/8.9 ˂
(X- µ)/ ˂
(47-42.3)/8.9)
Since Z = (x - µ)/
(40-42.3)/8.9=0.26
And (47-42.3)/8.9
= 0.53
Then P (40 ˂ X ˂ 47) = P (-0.26 ˂ Z ˂ 0.53)
P (-0.26 ˂ Z ˂ 0.53)
= 0.3045
% of students between 41 and 46 (inclusive) = 30.45%
Out of the 200 students in the Excel table, there are 37 who
are between 41-46 years (inclusive), giving an actual value of
37/200 X 100 = 18.5%
This is less than the computed 30.45%.
d) The chart below reflects
the age range of the students. It shows that the oldest 10% students are
between 59 years and 69 years.
This is as opposed to the data in the Excel table, where the
age range of the oldest 10% of the students is between 54 years and 59 years.
Question 2
Average
GPA of MBA students (µ) = ∑x/n
=
684.29/200 = 3.421
Standard deviation from average () = square root of ∑(x - µ)2/n
=
square root of ∑(0.0946 - 3.421)2/200
=
0.308
Margin
of error = Za/2 x /square root of n
= -1.96 x 0.308/14.14
= ± 0.043
95%
Confidence interval = 3.421 ± 0.043
Upper
confidence interval = 3.464
Lower
confidence interval = 3.378
This is
shown in the chart below:
The chart further has a probability of about 2.5% beyond the 4.0, which is
not actually possible. This is because of the 95% confidence interval.
Further, the lowest GPA in the chart is 2.4, which is lower than the lowest
GPA in the sample data on the Excel file of 2.8.
Question 3
Average
(µ) = ∑x/n
Where:
Finance=0
Leadership=1
Marketing=2
No Major=3
µ =
270/200 = 1.35
Standard
deviation from average () =
square root of ∑(x- µ)2/n
=
square root of ∑(1.284 – 1.35)2/200
=
1.133
Proportion of students majoring in
Finance is:
P(0 ˂ X ˂ 1) = P(0-1.35
˂X - µ ˂ 1-1.35)
P((0-1.35)/1.133 ˂
(X- µ)/ ˂ (1-1.35)/1.133)
Since Z = (x - µ)/
(0-1.35)/1.133=-1.19
And (1-1.35)/1.133
= 0.31
Then P (0 ˂ X ˂ 1) = P (-1.19 ˂ Z ˂ 0.31)
P (-1.19 ˂ Z ˂ 0.31) = 0.2613
Students majoring in Finance are represented between 0 and 1
on the x-axis in the chart below.
Margin
of error =
Za/2 x /square root of n
= -1.96 x 1.133/14.14
= ± 0.157
95%
confidence interval = 1.133
± 0.032
Upper
confidence interval = 1.165
Lower
confidence interval = 1.101
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