Question

100 POINTS! ANSWER NOW! I'm really slow so please explain this as thoroughly as possible.

a) The average height of sunflowers in a field is 64 in. with a standard deviation of 3.5 in. On a piece of paper, draw a normal curve for 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 came up with each of your labels.

b) If there are 3,000 plants in the field, approximately how many will be taller than 71 in.? Explain how you got your answer.

(For part b, when solving it out, please explain HOW you found out there is 2.5% taller than 71 inches, and do NOT copy and paste someone else's answer because I've already seen them all, and I will delete your answer.)

Answer 1

Explanation:

a) The graph of a normal curve with a mean of 64" and a standard deviation of 3.5" is a bell-shaped curve with its

maximum value at x = 64.

Important points:

mean - 3 deviations: 64 - 3(3.5) = 64 - 10.5 = 53.5

mean - 2 deviations: 64 - 2(3.5) = 64 - 7 = 57

mean - 1 deviation: 64 - 3.5 = 60.5

mean: = 64

mean + 1 deviation: 64 + 3.5 = 67.5

mean + 2 deviations: 64 + 2(3.5) = 64 + 7 = 71

mean + 3 deviations: 64 + 3(3.5) = 64 + 10.5 = 74.5

b) If there are 3000 plants, approximately how many will be taller than 71"?

71" is 2 standard deviations above the mean. The region between two standard deviations below the mean and two standard deviations holds approximately 95% of the plants.

Therefore there will be approximately 5% of the plants divided equally between being taller than two standard

deviations above the mean and being shorter than two standard deviations below the mean. Thus, there will be

approximately 2.5% of the plants taller than 71": 0.025 x 3000 = 75 plants.

or

the z-score for 71

= (71-64)/35 = 2

going to your tables, or some other suitable source for standard deviation, we find

P(z < 2) = .9772

P(z > 2) = .0228

.0228(3000) = 68.4

or appr 68 plants