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Assume that the population of diastolic measurement for blood pressure is normally distributed with a mean of 72 and a standard deviation of 12. What proportion of the population would have a diastolic measurement above 93. Draw the Normal distribution, and shade the region that correspond the proportion.

Question

Assume that the population of diastolic measurement for blood pressure is normally distributed with a mean of 72 and a standard deviation of 12.

What proportion of the population would have a diastolic measurement above 93. Draw the Normal distribution, and shade the region that correspond the proportion.

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Solution

To answer this question, we will use the concept of Z-score in statistics. The Z-score is a measure of how many standard deviations an element is from the mean.

Step 1: Calculate the Z-score The Z-score is calculated using the formula:

Z = (X - μ) / σ

where: X is the value from the dataset (in this case, 93) μ is the mean of the population (in this case, 72) σ is the standard deviation of the population (in this case, 12)

So, Z = (93 - 72) / 12 = 1.75

Step 2: Find the proportion corresponding to the Z-score We can find this value from the Z-table or use a statistical calculator. The value in the Z-table that corresponds to 1.75 is 0.9599. This means that 95.99% of the data falls below a diastolic measurement of 93.

Step 3: Find the proportion above 93 Since we know the proportion below 93, we can subtract this from 1 to find the proportion above 93.

So, 1 - 0.9599 = 0.0401

Therefore, approximately 4.01% of the population would have a diastolic measurement above 93.

Step 4: Draw the Normal distribution To draw the Normal distribution, you would plot a bell curve with the mean (72) at the center. The standard deviations would be marked along the x-axis. The area to the right of 93 (representing the Z-score of 1.75) would be shaded to represent the proportion of the population with a diastolic measurement above 93. This shaded area represents approximately 4.01% of the total area under the curve.

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