Answer: 169.2 per acre are produced if a field is in the top 25%.
Step-by-step explanation:
Given: The mean rain-fed corn yield of harvested fields are normally distributed, produce 147.2 bushels per acre with a standard deviation of 33 bushels per acre.
Let X denote the mean rain-fed corn yield of harvested fields.
Let x denotes the number of bushels per acre are produced if a field is in the top 25%.
Top 25% on a density curve means , (100-25%)i.e. 75% area lies below x.
I.e.
[tex]P(X<x)=P(\dfrac{X-\mu}{\sigma}<\dfrac{x-147.2}{33})=0.75\\\\\Rightarrow\ P(Z<\dfrac{x-147.2}{33})=0.75\ \ [Z=\dfrac{X-\mu}{\sigma}][/tex].
Z-score for p-value of 0.75 (one tailed)=0.67 [By z-table]
Then, [tex]\dfrac{x-147.2}{33}=0.6744[/tex]
[tex]\Rightarrow\ {x-147.2}=33\times0.67\\\\\Rightarrow\ {x-147.2}=22.11\\\\\Rightarrow\ {x}=22.11+147.2=169.31[/tex]
Here approx value is 169.2 (from options).
So, 169.2 per acre are produced if a field is in the top 25%.
