Respuesta :
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:
[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 0.90
P(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.
