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

ROLCAM · 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 · 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 $z = \frac{x - \mu}{\sigma}$ 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 · Oct 15, 2009, 11:15 AM

Do we do these in the S2 paper?

Chris-infj · Oct 16, 2009, 07:07 AM

Quote:

Originally Posted by Unknown008

Do we do these in the S2 paper?

S1 paper

Unknown008 · Oct 17, 2009, 08:15 AM

Ok...

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

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