Answer:
There is a 5.59% probability that the farmer will count 60 or more flights on average in the next 40 rows down which he drives his tractor.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
In this problem, we have that:
After several rows he figures the mean number of flights to be 57 with a standard deviation of 12, so
.
We have a sample of 40 rows, so we have to find the standard deviation of the sample to use in the place of
in the Z score formula.
![s = \frac{12}{\sqrt{40}} =1.897](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B12%7D%7B%5Csqrt%7B40%7D%7D%20%3D1.897)
What is the probability of the farmer will count 60 or more flights on average in the next 40 rows down which he drives his tractor?
This is 1 subtracted by the pvalue of Z when
.
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{60 - 57}{1.89}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B60%20-%2057%7D%7B1.89%7D)
![Z = 1.59](https://tex.z-dn.net/?f=Z%20%3D%201.59)
has a pvalue of 0.9441.
This means that there is a 1-0.9441 = 0.0559 = 5.59% probability that the farmer will count 60 or more flights on average in the next 40 rows down which he drives his tractor.