The average height of sunflowers in a field is 64 inches. With a standard deviation of 3.5 inches on a apiece of paper draw a normal curve from the distribution including the values on the horizontal axis at one two and three standard deviations from the mean describe your drawing in as much detail as possible and explain how you can came up with each of your labels
If there are 3000 plants in the field approximately much will be taller than 71 inches
My drawing would be a bell-shaped curve with the mean of 64 inches at the center. The horizontal axis would be labeled with the values at one, two, and three standard deviations from the mean. The value at one standard deviation from the mean would be 60.5 inches, two standard deviations from the mean would be 57 inches, and three standard deviations from the mean would be 53.5 inches.
To answer the question, approximately 600 plants will be taller than 71 inches. This can be calculated by using the 68-95-99.7 rule, which states that 68% of the data will be within one standard deviation of the mean, 95% of the data will be within two standard deviations of the mean, and 99.7% of the data will be within three standard deviations of the mean. Since 3000 plants are in the field, 68% of them (2040 plants) will be within one standard deviation of the mean (60.5-67.5 inches). Since 71 inches is greater than 67.5 inches, the remaining 960 plants (3000-2040) will be taller than 71 inches.
Looks like the bot doesn't know how the 68-95-99.7 rule works.
z-score for 71 = (71-64)/3.5 = 2
So appr 5% would be > 71 or < 57, or 2.5% would be > 71
If the bot had used this method it should have had the answer of .25(3000) or 75 trees, but it had it at 960
The proper way would be to use the normal distribution tables, z-scores etc.
I use an on-line website at "davidlane,com", (replace the , with a .)where you can enter the data
directly
I got prob( >71) = .0228
so of 3000 sunflowers, .0228(3000) or appr 68 would be over 71 inches tall