With food prices becoming a great issue in the world; wheat yields are even more important. Some of the highest yielding dry land wheat yields are in Washington state and Idaho; with an average of 100 bushels per acre and a known standard deviation of 30 bushels per acre. A seed producer would like to save the seeds from the highest 90% of their plantings. Above what yield should they save the seed (bushels/acre)?80.8138.478.473.061.6

Answer:Option E) 61.6Step-by-step explanation:We are given the following information in the question: Mean, μ = 100 bushels per acreStandard Deviation, σ = 30 bushels per acreWe assume that the distribution of yield is a bell shaped distribution that is a normal distribution. Formula: [tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]P(X>x) = 0.90 We have to find the value of x such that the probability is 0.90P(X > x)  [tex]P( X > x) = P( z > \displaystyle\frac{x - 100}{30})=0.90[/tex]  [tex]= 1 -P( z \leq \displaystyle\frac{x - 100}{30})=0.90 [/tex]  [tex]=P( z \leq \displaystyle\frac{x - 100}{30})=0.10 [/tex]  Calculation the value from standard normal table, we have,  [tex]P(z<-1.282) = 0.10[/tex][tex]\displaystyle\frac{x - 100}{30} = -1.282\\x = 61.55 \approx 61.6[/tex]  Hence, the yield of 61.6 bushels per acre or more would save the seed.