Probability of 20ml of rainfall when 13ml chance is 87.7% and 16.9ml chance is 40.9%

Farmers in District D know when to plant and harvest because rainfall follows a particular pattern which repeats itself every seven months. Weather records for this particular area indicate a normal distribution as far as rainfall is concerned. The records further reveal that the chances of at least 13.4 millimeters of rainfall have remained at 87.7%, while the chances of at most 16.9 millimeters remain at 40.9%. For the farmers to plant wheat they need at least 20 millimeters of rainfall, with what probability do they plant the wheat? SHOW ME HOW TO WORK THIS OUT! THANK YOU!

 ROLCAM Posts: 1,437, Reputation: 250 Ultra Member #2 Oct 15, 2009, 05:06 AM
This works out as follows:-
(by extrapolation in reverse).
(based on the heading details which varies from the details in the bulk of the question)

Number

1 13.0 87.7
2 13.1 86.5
3 13.2 85.3
4 13.3 84.1
5 13.4 82.9
6 13.5 81.7
7 13.6 80.5
8 13.7 79.3
9 13.8 78.1
10 13.9 76.9
11 14.0 75.7
12 14.1 74.5
13 14.2 73.3
14 14.3 72.1
15 14.4 70.9
16 14.5 69.7
17 14.6 68.5
18 14.7 67.3
19 14.8 66.1
20 14.9 64.9
21 15.0 63.7
22 15.1 62.5
23 15.2 61.3
24 15.3 60.1
25 15.4 58.9
26 15.5 57.7
27 15.6 56.5
28 15.7 55.3
29 15.8 54.1
30 15.9 52.9
31 16.0 51.7
32 16.1 50.5
33 16.2 49.3
34 16.3 48.1
35 16.4 46.9
36 16.5 45.7
37 16.6 44.5
38 16.7 43.3
39 16.8 42.1
40 16.9 40.9
41 17.0 39.7
42 17.1 38.5
43 17.2 37.3
44 17.3 36.1
45 17.4 34.9
46 17.5 33.7
47 17.6 32.5
48 17.7 31.3
49 17.8 30.1
50 17.9 28.9
51 18.0 27.7
52 18.1 26.5
53 18.2 25.3
54 18.3 24.1
55 18.4 22.9
56 18.5 21.7
57 18.6 20.5
58 18.7 19.3
59 18.8 18.1
60 18.9 16.9
61 19.0 15.7
62 19.1 14.5
63 19.2 13.3
64 19.3 12.1
65 19.4 10.9
66 19.5 9.7
67 19.6 8.5
68 19.7 7.3
69 19.8 6.1
70 19.9 4.9
71 20.0 3.7

Conclusion a 20 millimeters has a 3.7%
Probability.
 Chris-infj Posts: 31, Reputation: 15 Junior Member #3 Oct 15, 2009, 08:20 AM
Let X be the amount of rainfall in mm

It is given that X follows a normal distribution. What is not known is the mean and the standard deviation of rainfall.

You're given $P(X \geq 13.4mm)$ = 0.877.

1. Draw a bell-shaped curve and illustrate that.

2. Then look up in a standard normal table and find the z value that corresponds to a p value of 0.877.

3. Is it a positive z or a negative value? You'll know when you look at your drawing. In fact, it is strongly advised to draw a picture.

4. So then you use the formula $\Large z = \frac{x - \mu}{\sigma}$ to obtain a first equation involving $\mu$ and $\sigma$.

Repeat as from step 1. For the other piece of information $P(X \leq16.9)$ = 0.409.

Draw a picture for that too. Look up the z value corresponding to a p value of 0.409 in normal tables or use Excel Norminv function.

You should get two simultaneous equations with $\mu$ and $\sigma$ as two unknowns and solve to get them.

Then armed with these values, you can calculate $P(X \geq 20mm )$ .
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #4 Oct 15, 2009, 11:15 AM
Do we do these in the S2 paper?
 Chris-infj Posts: 31, Reputation: 15 Junior Member #5 Oct 16, 2009, 07:07 AM
Quote:
 Originally Posted by Unknown008 Do we do these in the S2 paper?
S1 paper
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #6 Oct 17, 2009, 08:15 AM
Ok...

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