Question

The mean yield for one acre plot is 662 kg with a s.d. 32 kg. 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

yield for one acre plot = random variable x
μ = mean of the distribution for yield = E (X) = 662 kg
standard deviation = σ = 32 kg

Normal distribution variable = X = (x - μ)/σ
cumulative probability function of the normal distribution:
            F(X ) = P (X <= x)

Probability p of yield x being in the range 600kg and 750kg = P(600 ≤ x ≤ 750)

  p   = P[ (600 - 662)/32  ≤ X  ≤ (750 - 662)/32 ]
        = P [ -1.9375 ≤ X ≤ 2.75 ]
        = P( X ≤ 2.75) - [ 1 - F( X ≤ 1.9375) ]

  =>  p  = F (2.75)  + F (1.9375) - 1

  Look up these values  in  normal distribution function tables.   I have the two values as below:

           p = 0.997  + 0.973 - 1  =  0.97  approximately.

You need to know how to read and interpret the values in the table.
 
0.97  :     This is the probability with which a randomly selected plot may produce a yield in the given range.

Number of plots we have =  N = 1000 plots    =  data  sample size.
 
Number of plots, that are likely to produce yield in the given range
        = N * p
       = 1000 * 0.97 =  970 plots.