# Thread: Probability and standard deviation

1. ## Probability and standard deviation

Dear Members,

I am struggling with this particular problem and I am not sure about my answers. If anybody could verify them that would be of great help. (I will post my answers below) This is not a homework problem. I am helping my niece solve practice problems and we are stuck

Historical data indicates that the starting salary for a new MBA graduate in a leading management consulting firm can be modeled as a normal distribution with mean $90,000 and standard deviation$20,000. Second year salaries increase by 20%. The bonus each year, for the first and following years, can be modeled as a normal distribution with mean $25,000 and standard deviation$5,000. We assume that the bonus is independent of the initial salary increase. For the purpose of this problem annual compensation is salary plus bonus.
(a) What is the expected annual compensation for a new hire?
(b) What is the standard deviation of the annual compensation for a new hire?
(c) What is the expected annual compensation after completing a year in the firm, I.e., just after the salary increase is announced?
(d) What is the probability that the annual compensation after completing a year in the firm, i.e., just after the salary increase is announced, will exceed \$140,000?
(Hint: The annual compensation is also normally distributed).

The answers I am getting are as follows:

(a) : 115,000
(b) : Not sure (is it just plain sum of standard deviations?)
(c) : 133,000
(d) : Stuck

For (d) I can imagine that I need to calculate the Z score but I am not able to structure the solution.

Thanks,

KRS

2. ## Re: Probability and standard deviation

a and c are correct

b) If X and Y are independent random variables that are normally distributed (and therefore also jointly so), then their sum is also normally distributed. link

X has mean 90,000 and sd = 20,000
Y has mean 25,000 and sd = 5,000

Z = (X + Y) has mean 90,000 + 20,000 and

Once you have the mean and sd you should be able to do part d

3. ## Re: Probability and standard deviation

Originally Posted by asterisk
a and c are correct

Once you have the mean and sd you should be able to do part d
Thanks Asterix! Appreciate the response.

BUT for (d) you will need to calculate the standard deviation of compensation after one year i.e after the salary has increased. The 20% salary increase has to feature somewhere right?

Won't the above SD (that you have calculated) hold true only while the candidate is in the first year? After completing a year an additional factors are kicking - 20% rise in salary.

That is something I am finding difficult to tackle.

Thanks,

KRS

4. ## Re: Probability and standard deviation

Yeah, you're right I missed that.

SO for a normal random variable X

E[c⋅X]=c⋅E[X]
Var[c⋅X]=c^2⋅Var[X]
sd[c⋅X]=c⋅sd[X]

Z = 1.2 * X + Y

mean = (1.2 * 90,000 +25,000)

5. ## Re: Probability and standard deviation

Can someone help me with this for a sample of n=75 the probability of a sample mean being greater than 221if u (mean) = 220 and o(standard deviation) =6.1 is ___. someone help me with this it is asking for the probability