# Thread: Calculating Probability from the Mean, and Standard Deviation

1. ## Calculating Probability from the Mean, and Standard Deviation

If anyone may be able to give me a little insight into how I would go about answering this question I would greatly appreciate it.

This is the question in its context,

A student obtains ANNUAL rainfall totals for Bedford, Bedforshire for the years 1900-1988. The student draws a histogram and finds that the station's annual rainfall totals can be approximated by a normal distribution with a mean of 752.4mm and a standard deviation of 105.578mm.
Calculate the probability that any given year will have a rainfall total of between 700-800mm.

Thank you in advance for any feedback.

2. Originally Posted by aunistudent
If anyone may be able to give me a little insight into how I would go about answering this question I would greatly appreciate it.

This is the question in its context,

A student obtains ANNUAL rainfall totals for Bedford, Bedforshire for the years 1900-1988. The student draws a histogram and finds that the station's annual rainfall totals can be approximated by a normal distribution with a mean of 752.4mm and a standard deviation of 105.578mm.
Calculate the probability that any given year will have a rainfall total of between 700-800mm.

Thank you in advance for any feedback.

You need to compute two Z-scores:

Z1= (700 - 752.4) / 105.578

Z2 = (800 - 752.4) / 105.578

Now, using your Z table in your textbook, find the proportion under the curve between Z1 and Z2 . That proportion will be your probability.

3. Thank you for your help!

This is my working,

Z1 = (700 – 752.4) / 105.578 = - 0.4963, -0.50 (2d.p)
= 0.1915 (Z value from table)

Z2 = (800 – 752.4)/ 105.578 = 0.4509, 0.45 (2d.p)
= 0.1736 (Z value from table)

Probability that any given year will have a rainfall total between 700 - 800mm = Z1 – Z2 = 0.1915 - 0.1737 = 0.0179

4. Originally Posted by aunistudent

This is my working,

Z1 = (700 – 752.4) / 105.578 = - 0.4963, -0.50 (2d.p)
= 0.1915 (Z value from table)

Z2 = (800 – 752.4)/ 105.578 = 0.4509, 0.45 (2d.p)
= 0.1736 (Z value from table)

Probability that any given year will have a rainfall total between 700 - 800mm = Z1 – Z2 = 0.1915 - 0.1737 = 0.0179

Well, your Z values are correct. But your probability that you computed is not correct. Keep trying......As a check, you should get a probability of about 0.365.

5. So you have to add Z1, and Z2 to get the 'whole' probability?
May I ask what the reasoning is behind this?

Thanks again,

6. Originally Posted by aunistudent
So you have to add Z1, and Z2 to get the 'whole' probability?
May I ask what the reasoning is behind this?

Thanks again,
No, you neither add nor substract the Z-Scores.

Rather, you have to use your Z-table to get the associated probabilites. Like this:

Prob{Z< 0.45} = 0.673645

Prob{Z< -0.50}= 0.308538

And, so, the probability of obtaining a value of Z between Z1 and Z2 is:

0.365 = 0.673645 - 0.308538

7. Originally Posted by Dragan
No, you neither add nor substract the Z-Scores.

Rather, you have to use your Z-table to get the associated probabilites. Like this:

Prob{Z< 0.45} = 0.673645

Prob{Z< -0.50}= 0.308538

And, so, the probability of obtaining a value of Z between Z1 and Z2 is:

0.365 = 0.673645 - 0.308538
I don't understand how you have got the following value,

Prob{Z< 0.45} = 0.673645

The values I read of the table for Z1 and Z2 were not cumulative, are they supposed to be?

8. Originally Posted by aunistudent
I don't understand how you have got the following value,

Prob{Z< 0.45} = 0.673645

The values I read of the table for Z1 and Z2 were not cumulative, are they supposed to be?
Your textbook (table) may not have a "larger portion" column like the one I am looking at right now.

You may just have a "Mean to Z" column where the value is .1736. So, the value of .6736 comes from adding 0.5 to 0.1736 to get the cumulative total.

Note: there's more than one way to arrive at the correct probability of .365.

9. okay, I understand now where you got the values from, it makes sense.

Thank you for all of your help!

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