may i know how to solve this question??

Question

A farmer will earn a profit of RM28,000 in case of heavy rain next year, RM55,000 in case of moderate rain, RM15,000 in case of little rain, and RM7,000 in case of no rain. A meteorologist forecasts that the probability is 0.30 for heavy rain, 0.35 for moderate rain, 0.20 for little rain, and 0.15 for no rain next year.

1. Construct the probability distribution and cumulative probability distribution of next year’s profit for this farmer.                                                                        (5 marks)

 

1. Calculate the expected of next year’s profit in thousands of ringgits for this farmer. Give a brief interpretation of the value obtained.                                                  (4 marks)

                             

1. Calculate the standard deviation of next year’s profit in thousands of ringgits for this farmer.                                                                                                                       (4 marks)

                                                                                                                                                                             

1. Discuss the meaning of standard deviation of the random variable focusing on the values obtained in (c) and (d) above.                                                                                                 (5 marks)

Answer 1

a) The table below shows the probability distribution, where P is probability:

RAINFALL	P	Profit (RM)
Heavy	0.3	28000
Moderate	0.35	55000
Little	0.2	15000
None	0.15	7000

 

Consider how the meteorologist got his figures. Let's say he measured rainfall over a period of 365 days and discovered the following:

No rainfall for 54.75 days

Little rainfall for 73 days

Moderate rainfall for 127.75 days

Heavy rainfall for 109.5 days

These periods of time would give the probabilities quoted.

Based on these, and using a prorata value for the farmer's profit, we can calculate his profit for these periods:

54.75/365*7000+73/365*15000+127.75/365*55000+109.5/365*28000=31700. So the answer to b) (your second a) is RM31700.

No rainfall for the whole year produces a profit of RM7000; but little or no rainfall requires adjustment. The time period is 54.75+73=127.75 which is 0.35 year. So the profit has to be extrapolated based on the factor 1/0.35. So we have (54.75/365*7000+73/365*15000)/0.35=RM11571.43. For rainfall ranging from none to moderate, the time period is 0.7 year. The adjustment or extrapolation factor is 1/0.7. This gives us 23300/0.7=RM33285.71. And, of course, all types of rainfall give us RM31700. The tables below show cumulative probability.

Working from heavy to no rainfall:

RAINFALL	P	Profit (RM)
Heavy	0.3	28000
Heavy/moderate	0.65	42538.46
Heavy/moderate/little	0.85	36058.82
All	1	31700

And from no to heavy rainfall:

RAINFALL	P	Profit (RM)
None	0.15	7000
None/little	0.35	11571.43
None/little/moderate	0.7	33285.71
All	1	31700

In both the last two tables we have datasets for which we can calculate mean and SD for 7 rainfall patterns:

c) and d) No rainfall at all; little or no rain; all but heavy rain; a mixture of all types; heavy to moderate; some daily rain; only heavy rain. Using figures from the tables we arrive at a mean of RM27164.92 and a SD of RM12094.61, which gives a range for the profit of RM15070.31 to RM39259.53. (SD is square root of VARIANCE. Both mean and variance are calculated by dividing the sums of the relevant columns by 7, the number of rainfall types.)

RAINFALL	Profit	Profit-mean	(Profit-mean)^2
None	7000	-20164.92	406624000
Little or none	11571.43	-15593.49	243156930
No heavy rain	33285.71	6120.79	37464070
Mixed	31700	4535.08	20566951
Heavy or moderate	42538.46	15373.54	236345732
Some daily rain	36058.82	8893.9	79101457
Only heavy rain	28000	835.08	697359
MEAN:	27164.92	VARIANCE:	146279500