Unformatted text preview: (6.) (5 points) Which data are more variable: a collection of tree diameters with a mean of 12.7
inches and a variance of 9.9 inches2 or a collection of mole densities with a mean of 8.7 moles
per acre and a standard deviation of 4.4 moles per acre? (Justify your answer, no guessing
allowed) Since the units of measure dzﬂer between the two datasets, we must calculate the caeﬁicients of
variation (C Vs) C V = standard deviation divided by the mean times [00% M 12.7 x100% : 24.78% C Vfor the diameters equals C V for the moles per acre equals %x 100% = 50.57% Therefore, the moles per acre are more variable (7.) (4 points) When taking satellite imagery of a forest, there is a 70% chance the sky will be
deﬁnitely be clear enough (i.e. no clouds) for the sensor to capture an image of the forest and a
30% chance the sky might be clear enough (clouds are present) to capture an image of the forest. When the sky is clear enough for the imagery, there is a 98% chance the sensor will function
properly. When clouds are present though, there is a 20% chance the sensor will function properly. .
What is the probability the sensor will ﬁmction properly?
P(sensorﬁmctions) : P(sensorﬁ4nctions and clear sky) + P(sensorﬁmctians and 01014051 sky) :[P(clear sky) 1 P(sensorﬁmctions given clear Sky)] + [P(elouds) x P(sensorﬁmclions given clouds) ] :[0 70 x 0.98] + [0.30 x 0.20] : 0.746 ...
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- Spring '07