probability of a future mean based on known mean and SD

Ares

New Member
#1
Okay, I've given this one a shot but I have no idea if I did it correctly. Any input would be much appreciated.

Here's the scenario:
"A salesperson sells an average of 1120 products a week, with a standard deviation of 146. He will get a bonus if, over a four-week period, the average number he sells per week is more than 1200. What is the probability that he will get a bonus next month?"

So I assumed I needed to find P(X> 1200)
For the Z-score: (x - μ) ÷ σ = (1200 - 1120) ÷ 146 = 80 ÷ 146 = 0.55
From the z-table: z = 0.7088
So, P(X > 1200) = 1 – P(X ≤ 1200) = 1 – P(Z ≤ 0.55) = 1 – 0.7088 = 0.2912
Thus, a probability of 29%

However, does this equation work since the 1200 is an average and not a single data point? Does the 4 (for weeks) need to be considered here? This is what's really throwing me off, and Google has not been kind.
Thank you.
 

Ares

New Member
#3
Hello
Hi Ares,

How does the average distribute?
What is the standard deviation of the average?
Hello,
Thank you for the reply.
The only Standard Deviation I know is the 146.
As for how it distributes, I don't know, unfortunately. The only information I have is that in my first post. :(
 

obh

Active Member
#4
I suggest you will learn about the distribution of the average (find a nice video)
The average distributes also normally. With the same mean and σ(mean)=σ/sqrt(n)
 

Ares

New Member
#5
I've watched several videos and read multiple tutorials, but I don't understand how I can apply the 'distribution of a mean' here. All the formulae I've found require a 'sample mean,' whereas I have the mean of a whole population. Additionally, they seem to require data points, which I don't have.
Is the desired >1200 mean supposed be treated as a 'sample' or data point?
Or is the known population mean (1120) supposed to be considered a sample, when taken with the unknown future mean?
Or, am I supposed to invent random data points within the range of the standard deviation?
The z-transform etc from the first post is still the only thing I can find that seems to work, but as I said, I don't known if it's correct or needs to include the time-frame somehow.

Thank you again for your assistance.
 

obh

Active Member
#6
Hi Ares,

When you don't know the population parameters you use the sample parameters, it is the same idea.
When you use sample standard deviation you should use t distribution instead of normal.

For your question:
n=4.
σ(mean)=146/sqrt(4)
Instead of using the population's distribution as you did N(1120,146), please use the average's distribution: N(1120,146/√4)

You may calculate yourself the standard deviation of the average using:
1. for independent variables: (otherwise you should add the covariance...)
Var(X+Y)=Var(X)+Var(Y)
2. Var(aX+b)=a^2Var(X)
 
#7
I suspect that this is a Poisson distribution situation.
If sales average = 4480/4 weeks, what is the probability of 4800/4 weeks. The 146 doesn't help.
4800 = 4*1200 or 3*1600 and 1*0, or a lotta others.
My computer surrenders >1000
A ?poor? guess:
Assume that the distribution is Normal, mu = 4480, CV is constant, sigma = 4*146 = 584 , 4800-4480 =320, 320/584 = .55 = Z
P(Z>.55) = .291
"Poor" if Normal differs from Poisson, and don't know.
I'd use 29.11538%,baffle the reader with digits.