Given that systolic blood pressure (SBP) is nearly normally distributed, we can use properties of the normal distribution to answer these questions. Let’s define:

• $\mu = 118$ mmHg (mean)
• $\sigma = 26$ mmHg (standard deviation)

1. Probability of Stage 2 High Blood Pressure (SBP ≥ 160 mmHg):

We want to find $P(X \geq 160)$, where $X$ is the systolic blood pressure. First, we convert this to a standard normal variable:

$Z = \frac{X - \mu}{\sigma} = \frac{160 - 118}{26} = 1.615$

Using the standard normal table, or a calculator, we find the probability that a standard normal variable is greater than 1.615. The cumulative probability to the left of 1.615 is approximately 0.9466. Therefore, the probability of SBP ≥ 160 is:

$P(X \geq 160) = 1 - 0.9466 = 0.0534$

So, 5.34% of adults in the USA have Stage 2 high blood pressure.

2. Probability of Stage 1 High Blood Pressure (140 ≤ SBP < 160 mmHg):

We now need to calculate the probability that the SBP is between 140 and 160. We already know the $Z$-score for 160 is 1.615. Let’s calculate the $Z$-score for 140:

$Z = \frac{140 - 118}{26} = 0.846$

Using the standard normal table, the cumulative probability for $Z = 0.846$ is approximately 0.8013. We already found the cumulative probability for $Z = 1.615$ (160 mmHg) to be 0.9466. Therefore, the probability of Stage 1 high blood pressure is:

$P(140 \leq X < 160) = 0.9466 - 0.8013 = 0.1453$

So, 14.53% of adults in the USA have Stage 1 high blood pressure.

3. Systolic BP for the 30th Percentile:

The 30th percentile corresponds to a cumulative probability of 0.30. Using the standard normal table or a calculator, the $Z$-score corresponding to 0.30 is approximately $Z = -0.524$.

Now, we convert this $Z$-score back to the systolic blood pressure:

$X = Z\sigma + \mu = (-0.524)(26) + 118 = 104.38 \, \text{mmHg}$

So, if you are in the 30th percentile for blood pressure, your systolic BP is approximately 104.38 mmHg.

4. Systolic BP for the Top 2.5%:

The top 2.5% of adults corresponds to a cumulative probability of 0.975 (since we are looking for the 97.5th percentile). From the standard normal table, the $Z$-score corresponding to 0.975 is approximately $Z = 1.96$.

Converting this back to systolic blood pressure:

$X = Z\sigma + \mu = (1.96)(26) + 118 = 168.96 \, \text{mmHg}$

So, the systolic blood pressure that cuts off the top 2.5% of adults is approximately 169 mmHg.

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Summary:

1. Stage 2 High BP (SBP ≥ 160 mmHg): 5.34%
2. Stage 1 High BP (140 ≤ SBP < 160 mmHg): 14.53%
3. 30th Percentile Systolic BP: 104.38 mmHg
4. Systolic BP for Top 2.5%: 169 mmHg

Would you like any further explanation or details on these calculations?

Further Questions:

1. What is the probability that an adult has systolic BP less than 100 mmHg?
2. What percentile corresponds to a systolic BP of 130 mmHg?
3. What is the probability that an adult has systolic BP between 110 and 130 mmHg?
4. If an adult has systolic BP of 145 mmHg, what percentile are they in?
5. What percentage of adults have systolic BP below 90 mmHg?

Tip:

When dealing with normal distributions, understanding how to use $Z$-scores and cumulative probabilities helps in solving a wide range of probability-related problems.