Explain this on a ti 84 calculator please- A farmer was interest in determining

Question

Explain this on a ti 84 calculator please-

A farmer was interest in determining how many grasshoppers were in his field. He knows that the distibution of grasshoppers may not be normally distributed in his field due to growin conditions. As he drives his tracor down each row he counts how many grasshoppers he sees flying away. After several rows he figures the mean number of flights to be 57 with a standard deviation of 12. What is the probability that the farmer will count 52 or fewer flights on average in the next 40 rows down which he drives his tractor?

Explanation / Answer

NORMAL DISTRIBUTION
the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd/sqrt(n) ~ N(0,1)
mean ( u ) = 57
standard Deviation ( sd )= 12/ Sqrt ( 40 ) =1.8974
sample size (n) = 40

the probability that the farmer will count 52 or fewer flights on average in the next 40 rows down which he drives his tractor
P(X > 52) = (52-57)/12/ Sqrt ( 40 )
= -5/1.897= -2.6352
= P ( Z >-2.6352) From Standard Normal Table
= 0.9958,
P(X < = 52) = (1 - P(X > 52)
= 1 - 0.9958 = 0.0042

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