probability of a future mean based on known mean and SD


New Member
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.


New Member
Hi Ares,

How does the average distribute?
What is the standard deviation of the average?
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. :(


Active Member
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)


New Member
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.


Active Member
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:
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...)
2. Var(aX+b)=a^2Var(X)
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.