Mean, Standard Deviation & Confidence Intervals

Computing Assignment
Bus105
Student id
[Pick the date]

Section 1
Data
a) Scatter plot highlighted below:
The variables income and annual contribution is showing linear positive relationship with
strength of moderate to strong.
b) Regression line from scatter plot

y=0.1197+1829.1
y represents: annual contribution
x represent: Income
Annual contribution needs to be determined for income x = $200,000
y = (0.1197*200000) +1829.1 = $25769.1
Therefore, the value annual contribution for given income ($200,000) is $25769.1.
c) The z score needs to be determined.
Given average of the estimates (μ) = $27,000
Standard deviation (σ ) = $2,100
x ( above ) =$ 25769.1
Formula for z score : z={ x−μ
(σ) }
z score=25769.1−27000
2100
Vaue of z score=−0.58614
Hence, z score=−0.58614 −0.59
d) Value of P (Z < Z score) needs to be determined.
Value would be computed by using excel function “NORMSDIST () ”
P (Z < Z score) = P (Z<−0.58614 ¿

¿ NORMSDIST ( −0.58614 )
¿ 0.27889
e) Value of estimated rank would be shown below:
Value of estimated rank = {P (Z < Z score)}*10000
¿ P ( Z ←0.58614 )∗10000
¿ ( 0.27889 ) ∗10000
¿ 2788.9 2789
Section 2
a) The first pivot table indicates high risk investment (riskier) and loss and profit relation.
The respective sample size for riskier type n1=81
Proportion for riskier investment that would create a loss P1= 12
81 =0.1481
The second pivot table indicates low risk investment (safer) and loss and profit relation.