Calculate Wheat Field Yields: Mean of 10 Tons/Acre with 2 SD - Practice Problem

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The discussion revolves around calculating the number of wheat fields yielding between 8 to 12 tons per acre, given a mean of 10 tons and a standard deviation of 2 tons. It clarifies that 68.3% of the data falls within one standard deviation of the mean, which applies to this scenario. Therefore, approximately 68 fields out of 100 should yield within that range. If the range were extended to 6 to 14 tons, 95 fields would be expected to fall within that two-standard-deviation interval. The key takeaway is understanding the difference between one and two standard deviations in normal distribution calculations.
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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.
 
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I'm hoping the answer is 68. I don't understand at what point you differentiate betwen multiplying by 95.4% or 68.3%.
 
taffyleg said:
I'm hoping the answer is 68. I don't understand at what point you differentiate betwen multiplying by 95.4% or 68.3%.

68.3% the the area under the normal curve within 1 standard deviation; 95.4% is the area under the normal curve within 2 standard deviations.

Since the question gives you 1 standard deviation in either direction, your answer is correct. 95 fields would be the correct answer if they asked for the numbr with 6 to 14 tons.
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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