Question

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)?
80.8

138.4

78.4

73.0

61.6

Answer 1

Answer:

Option E) 61.6

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 100 bushels per acre

Standard Deviation, σ = 30 bushels per acre

We assume that the distribution of yield is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(X>x) = 0.90

We have to find the value of x such that the probability is 0.90

P(X > x)  

$P( X > x) = P( z > \displaystyle\frac{x - 100}{30})=0.90$  

$= 1 -P( z \leq \displaystyle\frac{x - 100}{30})=0.90$  

$=P( z \leq \displaystyle\frac{x - 100}{30})=0.10$  

Calculation the value from standard normal table, we have,  

$P(z$

$\displaystyle\frac{x - 100}{30} = -1.282\\x = 61.55 \approx 61.6$  

Hence, the yield of 61.6 bushels per acre or more would save the seed.