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
a) The average height of sunflowers in a field is 64 inches with a standard deviation of 3.5 inches. Describe a normal curve for the distribution, including the values on the horizontal axis at one, two, and three standard deviations from the mean.
b) If there are 3,000 plants in the field, approximately how many will be taller than 71 inches?
9 answers
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.
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
see
http://davidmlane.com/normal.html
You can enter the data as is, click on above
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
see
http://davidmlane.com/normal.html
You can enter the data as is, click on above
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1 answer