An image illustrating the concept of normal distribution using a sunflower field. The average sunflower height is visually represented as being 64 inches, with taller and shorter sunflowers at varying distances from the mean, reflective of the standard deviations. Some unique aspects include sunflowers represented in varying heights of one, two, and three standard deviations. Additionally, visually exhibit a section of the field depicting around 3000 sunflowers, with some specific area highlighting sunflowers that are taller than 71 inches.

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?

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.

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a) To describe a normal curve for the distribution of sunflower heights in the field, we can use the concept of standard deviations from the mean. A normal distribution is symmetrical, bell-shaped, and often referred to as a bell curve.

In this case, the average height of the sunflowers is 64 inches with a standard deviation of 3.5 inches. The mean (average) is located at the center of the normal curve, and in this case, it is 64 inches.

To identify the values on the horizontal axis at one, two, and three standard deviations from the mean, we can use the following formula:

For one standard deviation:
Value = Mean + (1 * Standard Deviation)

For two standard deviations:
Value = Mean + (2 * Standard Deviation)

For three standard deviations:
Value = Mean + (3 * Standard Deviation)

Substituting the given values, we get:

For one standard deviation:
Value = 64 + (1 * 3.5) = 67.5 inches

For two standard deviations:
Value = 64 + (2 * 3.5) = 71 inches

For three standard deviations:
Value = 64 + (3 * 3.5) = 74.5 inches

These values represent the distances from the mean at which we can expect the majority of sunflower heights to fall.

b) In order to approximate the number of plants taller than 71 inches, we need to consider the proportion of plants that would fall within that range.

To do this, we can use a concept known as the empirical rule (also called the 68-95-99.7 rule) for a normal distribution. According to this rule:

- Approximately 68% of the data falls within one standard deviation from the mean.
- Approximately 95% of the data falls within two standard deviations from the mean.
- Approximately 99.7% of the data falls within three standard deviations from the mean.

Since we know the mean and the standard deviation, we can calculate the percentage of sunflowers taller than 71 inches by finding the area under the normal curve above that point.

Using a statistics software or a standard normal distribution table, we can find the percentage (probability) associated with the z-score corresponding to a height of 71 inches.

Once we obtain the percentage, we can estimate the number of plants taller than 71 inches by multiplying it by the total number of plants in the field, which is given as 3,000.

Please note that the exact value can vary depending on the shape of the distribution and the accuracy of the approximation method used.

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

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