Results Displayed in a Histogram:
The histogram has a distinct bell-shaped curve which proves that the weights follow a normal distribution, which now means I have to calculate the mean and the standard deviations of the weights.
Process
Add up all 80 numbers previously listed above and this leads to the final total weight= 6983
Average the values to find the mean weight.
6983/80 ≈ 87.2875
Next, find the upper quartile(UQ) and lower quartile(LQ) to find the variance
LQ= 64 g
UQ= 110 g
IQR (variance)=46 g
The Standard Deviation
√46 ≈ 6.78 g
In general, there should be about roughly ⅔ (68.3%) of the mass in one standard deviation of the mean, roughly 95.4% between two standard deviations of the mean, and 99.7% between three standard deviations of the mean. However, it is very rare that any value would be more than three standard deviations.
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51 g and 87.2875 + 6.78= 94.01 g
Between 80.51 g and 94.01 g there are exactly 37/80 x 100= 46.25%. This is under value for one standard deviation which leads to speculation about if there was a large harvest during the year creating large apples.
2 standard deviations from the mean is 87.2875 - (2 x 6.78)= 73. 73g and 87.625 + (2x 6.78)= 100.85 g.
Between 74.075 g and 101.195 g there are 62 apples which means 62/80 x 100= 77.5%. Again, this value is too low because between two standard deviations there should be 99.7% which leads myself to believe that there wasn’t enough apples collected or a data collection error.
3 standards of deviation from the mean is 87.2875- (3 x 6.78) = 66.95 g and
87.2875 + (3 x 6.78) = 107.63 g
Between 66.95 g and 107.63 g there are 77 apples within this range. This means 77/80 x 100= 96.25%. After discovering the percentage was undervalue seeing that at three derivatives it should be at 99.7%, I took the fourth derivative of the apples’