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.
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
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
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.
okay, I understand now where you got the values from, it makes sense.
Thank you for all of your help!
Tweet |