Respuesta :
Expected Returns:You have chosen biology as your college major because you wouldlike to be a medical doctor.However, you find that the probability of being accepted tomedical school is about 10 percent.If you are accepted to medical school, then yourstarting salary when you graduate will be $300,000 per year.However, if you are notaccepted, then you would choose to work in a zoo, where you will earn $40,000 per year.Without considering the additional educational years or the time value of money, what isyour expected starting salary as well as the standard deviation of that starting salary?Solution:E(salary) = 0.9($40,000) + (0.1) ($300,000) = $66,000σ2salary= 0.9($40,000 - $66,000)2+ (0.1) ($300,000 - $66,000)2=$6,084,000,000 =>σsalary= ($6,084,000,000)1/2= $78,0007.4Historical Market Performance:Describe the general relation between risk and returnthat we observe in the historical bond and stock market data.Solution:
The expected starting salary is $66,000 and the standard deviation of that starting salary is $235,440.
It is required to calculate the expected starting salary and standard deviation of that salary.
What is the standard deviation?
It is defined as the measure of data dispersement, It gives an idea about how much is the data spread out.
We have a probability of being accepted into medical school is P(a):
P(a) = 10% = 0.10.
Probability of being not accepted is P(r):
P(r) = 1 - P(a) ⇒ 1 - 0.10 ⇒ 0.90.
If, accepted into medical school, then salary S(a) = $300,000 per year.
If not accepted, then you would choose to work in a zoo, then salary:
S(r) = $40,000 per year.
Expected starting salary = P(a)×S(a) + P(r)×S(r)
= 0.10 × 300,000 + 0.90 × 40,000
= 30,000 + 36,000
= $66,000
We know the variance formula:
[tex]\rm V^2 =\frac{\sum(x_i-x')^2}{n-1}[/tex]
Where [tex]\rm V^2[/tex] is Variance, [tex]\rm x_i[/tex] is the value of one observation, [tex]\rm x'[/tex] is the mean value of all observations, and [tex]\rm n[/tex] is the number of observations.
Here in the question we have: x' = 66,000 and n = 2 sample.
[tex]\rm x_1[/tex] = 300,000 and [tex]\rm x_2[/tex] = 40,000 putting these values in the above formula:
[tex]\rm V^2 =\frac{\sum(300,000-66,000)^2}{2-1}+\frac{\sum(40,000-66,000)^2}{2-1}\\\rm V^2 = (234,000)^2+(-26,000)^2\\\rm V^2 = 55,432,000,000[/tex]
We know the:
[tex]\rm Standard \ deviation \ = \sqrt{Variance}[/tex]
[tex]\rm Standard \ deviation \ = \sqrt{55432000000}\\\rm Standard \ deviation \ = $235,440[/tex]
Thus, the expected starting salary is $66,000 and the standard deviation of that starting salary is $235,440.
Learn more about the standard deviation here:
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