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Question # V The set of eight random numbers shown depicts measured wind-speeds v (in mph) of a hurricane in eight directions at a location. The associated probability values p(v) of the RV are also listed. S E | 105 | 95 W | 110 NE | 102 NW | 98 SE | 90 SW | 115 | Wind directions N Wind-speed (v in | 100 | mph): RV Probability of 0.1995 occurrence: p (v) 0.03 0.058 0.068 0.0065 0.223 0.250 0.155 Determine: (i) Expected value of the wind-speed v (in mph) of the hurricane; (ii) decide the associated spread of the RV and justify/explain it; and, (iii) decide the wind direction, which is most likely. [Answer hints: E[v] =99.9 mph; mode = 90 mph]

Answer 1

Solution:

i. As, we know that, ΣυXpίυ) Expected value of wind speed, E(v)= L Value taken from given table in question. = 100 * 0.1995 + 105 * 0.03 +95 * 0.058 + 110 * 0.068 + 102 * 0.0065 +98 * 0.223 +90 * 0.25 + 115 * 0.155 Therefore, E(v) = 98.932 mph ii. E(v2) = v2 x p(u) = 1002 * 0.1995 + 1052 * 0.03 +952 * 0.058 + 1102 * 0.068 + 1022 * 0.0065 +982* 0.223 +902 * 0.25 + 1152 * 0.155 Therefore, E(v2) = 9956.193 Variance of the wind-speed, Var(v) = E(v2) - E(v)2= 9956.193 - 98.9322 = 168.6524 Associated spread of the RV =0=V168.6524 = 12.98662 We can see that all given values lie within 3 standard deviations from mean (98.932 -3 * 12.98662, 98.932 +3 * 12.98662) (59.97214, 137.8919) iii. The highest probability is 0.250 for the wind speed 90 mph SE. The wind direction which is most likely is SE.