The mean yield for one acre plot is 662 kilos with standard deviation 32 kilos assuming normal distribution, how many one acre plot in a batch of1000 plots would you expect to have yield (1) 700 kilos(2)below 650 kilos and (3) what is the lowest yield of the best 100 plots? Given [p<=z<=1.19=0.3830, p(0<=z<=0.38) =0. 1480, p(0<=z<=1.28) =0.4
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mean yield = m = 662 kg
standard deviation = 32 kg
for x=600 kg
z = 600 - 662/32 = -1.9375
for x=750 kg
z = 750-662/32 = 2.75
Using Z table we get the values;
= 0.0268 and 0.9970
so,
P(-1.9375 < z < 2.75)
= 0.9970 - 0.0268
= 0.9702
Therefore, the probability of land between 600-750 kg =0.9702
number of total plots = 1000
Number of plots with expected yield= N = 1000×0.9702 = 970.2
standard deviation = 32 kg
for x=600 kg
z = 600 - 662/32 = -1.9375
for x=750 kg
z = 750-662/32 = 2.75
Using Z table we get the values;
= 0.0268 and 0.9970
so,
P(-1.9375 < z < 2.75)
= 0.9970 - 0.0268
= 0.9702
Therefore, the probability of land between 600-750 kg =0.9702
number of total plots = 1000
Number of plots with expected yield= N = 1000×0.9702 = 970.2
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