Problem

Suppose that diastolic blood pressure readings of adult males have a bell-shaped distribution with a mean of $80 \mathrm{mmHg}$ and a standard deviation of $10 \mathrm{mmHg}$. Using the empirical rule, what percentage of adult males have diastolic blood pressure readings that are less than $70 \mathrm{mmHg}$ ? Please do not round your answer.

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Final Answer: The percentage of adult males who have diastolic blood pressure readings that are less than \(70 \mathrm{mmHg}\) is approximately \(\boxed{16\%}\).

Steps

Step 1 :The empirical rule, also known as the 68-95-99.7 rule, states that for a normal distribution, almost all data falls within three standard deviations of the mean. Specifically, 68% of data falls within one standard deviation, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

Step 2 :In this case, the mean is 80 mmHg and the standard deviation is 10 mmHg. A diastolic blood pressure reading of 70 mmHg is one standard deviation below the mean.

Step 3 :According to the empirical rule, approximately 50% of readings will be above the mean and 50% will be below. Since 68% of readings fall within one standard deviation of the mean, and we know that half of these will be below the mean, we can calculate that approximately 34% of readings will be less than 70 mmHg.

Step 4 :However, this is not the final answer. The 34% includes both the readings that are exactly at 70 mmHg and those that are below. We need to subtract the percentage of readings that are exactly at 70 mmHg.

Step 5 :Since the distribution is continuous, the probability of getting a reading exactly at 70 mmHg is theoretically 0. But in practice, we can consider it as half of the percentage of readings within one standard deviation from the mean, which is 68% / 2 = 34%.

Step 6 :So, the percentage of readings that are less than 70 mmHg is 34% - 34% / 2 = 16%.

Step 7 :Final Answer: The percentage of adult males who have diastolic blood pressure readings that are less than \(70 \mathrm{mmHg}\) is approximately \(\boxed{16\%}\).

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