Answer:
= 0.0041
Step-by-step explanation:
Given that:
A farmer was interest in determining how many grasshoppers were in his field. He knows that the distribution of grasshoppers may not be normally distributed in his field due to growing conditions. As he drives his tractor down each row he counts how many grasshoppers he sees flying away
mean number of flights to be 57
a standard deviation of 12
fewer flights on average in the next 40 rows
[tex]\mu = 57\\\\\sigma=12\\\\n=40[/tex]
so,
[tex]P(x<52)[/tex]
[tex]=P(\frac{x-\mu}{\sigma/\sqrt{n} } <\frac{52-57}{12/\sqrt{40} } )\\\\=P(z<\frac{-5\times6.325}{12} )\\\\=P(z<\frac{-31.625}{12})\\\\=P(z<-2.64)[/tex]
using z table
= 0.0041
The probability of the farmer will count 52 or fewer flights on average in the next 40 rows down which he drives his tractor is 0.0041 and this can be determined by using the properties of probability.
Given :
The probability of the farmer will count 52 or fewer flights on average in the next 40 rows down which he drives his tractor, can be determined by using the following calculations:
[tex]\rm P(x<52)=P\left (\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n} }}<\dfrac{52-57}{\dfrac{12}{\sqrt{40} }}\right)[/tex]
[tex]\rm P(x<52)=P\left (z<\dfrac{-5\times 6.325}{12 }}\right)[/tex]
[tex]\rm P(x<52)=P\left (z<\dfrac{-31.625}{12 }}\right)[/tex]
[tex]\rm P(x<52)=P\left (z<-2.64\right)[/tex]
Now, using z-table:
P(x < 52) = 0.0041
For more information, refer to the link given below:
https://brainly.com/question/21586810
