Question

the mean yield for one acre plot is 662 kg with a s.d 32kg .assuming normal distribution how many one acre plot in a batch of 1000 plots would your expect to have yield between 600 and 750 kg.

Answer 1

mean(m) = 662 kg
sd = 32 kg

for x=600 kg, z = $\frac{600-662}{32}=-1.9375$
for x=750 kg, z = $\frac{750-662}{32} =2.75$

Using Z table:
P(z < -1.9375) = 0.0268
P(z < 2.75) = 0.9970

P(-1.9375 < z < 2.75) = 0.9970 - 0.0268 = 0.9702

So probability of a land with yield between 600 kg and 750 kg is 0.9702
total plots = 1000
Number of plots with expected yield between 600 and 750 kg is given by
N = 1000×0.9702 = 970.2 ≈ 970