Question 1178064: Let X be a random variable that represents the diastolic blood pressure (DBP) of the population of 18- to 74-year-old men in the United States who are not taking any corrective medication. Suppose that X has mean 80.7 mm Hg and standard deviation 9.2.
(a) Obtain a bound on the probability that the DBP of this population will assumes values
between 53.1 and 108.3 mm Hg.
(b) In addition, assume that the distribution of X is approximately normal. Using the
normal table, find P(53.1 ≤ X ≤ 108.3). Compare this with the empirical rule.
Thank you :)
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**Given Information**
* X: Diastolic blood pressure (DBP)
* μ (mean) = 80.7 mm Hg
* σ (standard deviation) = 9.2 mm Hg
**(a) Chebyshev's Inequality**
Chebyshev's Inequality provides a bound on the probability that a random variable falls within a certain range. It states:
* P(|X - μ| ≥ kσ) ≤ 1/k²
* Or equivalently, P(|X - μ| < kσ) ≥ 1 - 1/k²
We want to find P(53.1 ≤ X ≤ 108.3). Let's rewrite this as:
* |X - μ| < kσ
* |X - 80.7| < k(9.2)
We need to find the range of X:
* 108.3 - 80.7 = 27.6
* 80.7 - 53.1 = 27.6
So, we have:
* |X - 80.7| < 27.6
Now, find k:
* k(9.2) = 27.6
* k = 27.6 / 9.2 = 3
Now, apply Chebyshev's Inequality:
* P(|X - 80.7| < 27.6) ≥ 1 - 1/k²
* P(53.1 ≤ X ≤ 108.3) ≥ 1 - 1/3²
* P(53.1 ≤ X ≤ 108.3) ≥ 1 - 1/9 = 8/9 ≈ 0.8889
Therefore, a bound on the probability is 8/9 or approximately 0.8889.
**(b) Normal Distribution and Empirical Rule**
Assume X is normally distributed.
1. **Calculate Z-scores:**
* Z1 = (53.1 - 80.7) / 9.2 = -27.6 / 9.2 = -3
* Z2 = (108.3 - 80.7) / 9.2 = 27.6 / 9.2 = 3
2. **Find P(-3 ≤ Z ≤ 3) using Normal Table:**
* P(-3 ≤ Z ≤ 3) = P(Z ≤ 3) - P(Z ≤ -3)
* From the normal table, P(Z ≤ 3) ≈ 0.9987 and P(Z ≤ -3) ≈ 0.0013
* P(-3 ≤ Z ≤ 3) = 0.9987 - 0.0013 = 0.9974
3. **Empirical Rule (68-95-99.7 Rule):**
* The empirical rule states that approximately 99.7% of the data falls within 3 standard deviations of the mean in a normal distribution.
* This is consistent with our calculated value of 0.9974.
**Comparison**
* **Chebyshev's Inequality:** Provides a lower bound (8/9 ≈ 0.8889). It is a general result and works for any distribution with a defined mean and standard deviation.
* **Normal Distribution:** Provides a more precise probability (0.9974) when the distribution is known to be normal.
* **Empirical Rule:** Agrees with the Normal Distribution calculation.
**Results**
(a) A bound on the probability is 8/9 or approximately 0.8889.
(b) P(53.1 ≤ X ≤ 108.3) ≈ 0.9974. This agrees with the empirical rule.
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