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# Change of earning \$380 a week

Ok based on the answer to the linked question, let me check I've understood this.

First a recap of the original question.

Each job pays \$10 on average with a sd of \$1.50.
I get a maximum of 70 jobs a week.
What is the probability I'll earn \$800 in a week?

Get the average income based on 70 jobs: 70 x \$10 = \$700.
Take the original sd 1.5 and square it: 2.25.
Multiply that by the number of jobs - 70 = 157.5.
Take the square root of that: \$12.55.

Now work out the size of the increase to make \$800: \$800 - \$700 = \$100.
Divide that my the new sd: 100/12.55 = 7.968
Try to look that up in a normal distribution and it is off the scale, so it is very unlikely you will earn \$800 in a week.

Now given that, here is a new question:
A job pays \$5.67 with an sd of \$1.96.
There is a maximum of 65 customers a week.
What is the probability of earning \$380.

65 x 5.67 = 368.55
1.96^2=3.8416 x 65 = 249.704
sqrt(249.704) = 15.802

380 - 368.55 = 11.45
11.45 / 15.802 = 0.7246

Look this up in the normal distribution and you get a probability of 76.42%.
This is the probability of making at least \$380 in a week.
Is this correct?
0
purplesoup
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1 Solution

Commented:
correct
-
here is a link to an easy to use t=calculator
http://www.mathsisfun.com/data/standard-normal-distribution-table.html
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Author Commented:
So the table you linked to is totally different to the table I used.

If you look at http://en.wikipedia.org/wiki/Standard_normal_table

You can see the table I used as the first one listed - the cumulative table, from this looking up 0.7246 (or actually 0.72) gives the answer I gave - 76.42%.

But under this is the cumulative from mean table which is the one you referenced. According to this table the reading for 26.424%.

Reading above in the article apparently the top table is used for less than percentages, and a complementary cumulative for greater than.

If this is the case perhaps I should calculate 1-0.7642 = 23.58%.

Also according to the wiki article the "Cumulative from mean gives a probability that a statistic is between 0 (mean) and Z. Example: Prob(0 = Z = 0.69) = 0.2549".

This doesn't seem to be what I'm looking for so are you sure you referenced the right table?

Can someone also tell me if I should be using the cumulative or complementary cumulative result?
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Commented:
The table I referenced gives the SAME number as you got, you just have to read it as it was designed..

You want the cumulative result.

Did you use from my link the table or the very revealing and well designed animated graph?

The graph (and the table) gives you the probability of the desired number being greater than the average so to the result from the graph (or the table) you have to add 50%.
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Author Commented:
Ok first off I assumed if I was right someone would just have said "yes, that's it", so if I get an answer other than that I'm assuming I'm wrong.

That said, looking at the initial figures I would have thought it was quite probable that you could make \$380, so I would guess the 76% was more likely than 25%, but I need to understand the maths.

This is the example your table gives:
Example: Percent of Population Between 0 and 0.45
Start at the row for 0.4, and read along until 0.45: there is the value 0.1736
So 0.1736 of the population are between 0 and 0.45 Standard Deviations from the Mean
And 0.1736 is 17.36%
So 17.36% of the population are between 0 and 0.45 Standard Deviations from the Mean

Erm basically I'm confused now - have I got a sd of 0.7246? or is it 1-0.7246 ??

This is the problem with trying to understand maths from the internet/wiki, it is always way too complicated, all I can find are questions and not answers and when I ask on EE all I get are ambiguous answers that seem to imply if I don't get half-written suggestions I'm an idiot.
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Commented:
note the first work in my answer. It was
"correct"
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