Calculating Wheat Field Yield Variance: A Quick Statistics Problem

[taffyleg]

For a sample of 100 wheat fields, the mean yield was 10 tons/acre and the standard deviation 2. Approximately how many fields out of the 100 should have yields in the 8 to 12 ton range? Show work.

 

[HallsofIvy]

But you haven't (shown your work, that is). What have you done on this? Do you know how to find the standard normal variable? Do you have a table of the normal probability distribution? If not, here's a good one on line:
http://people.hofstra.edu/Faculty/Stefan_Waner/RealWorld/normaltable.html

 

[Architeuthis Dux]

I would approach this problem by first understanding the context and assumptions of the data. It is important to note that the sample of 100 wheat fields may not be representative of the entire population of wheat fields. Additionally, the mean and standard deviation provide information about the central tendency and variability of the sample, but they do not account for the distribution of the data.

To calculate the number of fields with yields in the 8 to 12 ton range, we can use the empirical rule, also known as the 68-95-99.7 rule, which states that approximately 68% of the data falls within one standard deviation from the mean, 95% within two standard deviations, and 99.7% within three standard deviations.

In this case, we know that the mean yield is 10 tons/acre and the standard deviation is 2, which means that approximately 68% of the fields should have yields between 8 and 12 tons/acre. This can be calculated by first finding the z-scores for 8 and 12 tons/acre:

z-score for 8 tons/acre = (8 - 10)/2 = -1
z-score for 12 tons/acre = (12 - 10)/2 = 1

Using a standard normal table, we can find that the area under the curve between -1 and 1 is approximately 68%. This means that out of the 100 fields, approximately 68 will have yields between 8 and 12 tons/acre.

It is important to keep in mind that this is an estimate based on the assumption that the data follows a normal distribution. If the data is skewed or has outliers, this estimate may not be accurate. Additionally, as mentioned earlier, the sample may not be representative of the entire population of wheat fields. Therefore, further analysis and consideration of the context and assumptions of the data is necessary to provide a more accurate response.