Question

Suppose that the mean systolic blood pressure for women over age seventy is 131 mmHg (millimeters of mercury), with a standard deviation of 9mmHg. Suppose that the blood pressures are normally distributed. Complete the following statements (choose the correct percentage) 68%, 75%, 95%, 99.7%

Answer 1

Given:

Mean, μ = 131 mmHg

Standard deviation, σ = 9 mmHg

Let's solve for the following:

• (a). Approximately 99.7% of the women over seventy have blood pressures between?

Since it is a normal distribution, we know that 99.7% of data lies between:

μ - 3σ(3 standard deviations below the mean) and μ + 3σ (3 standard deviations above the mean)

Thus, we have:

μ - 3σ = 131 - 3(9) = 131 - 27 = 104 mmHg

μ + 3σ =131 + 3(9) = 131 + 27 = 158 mmHg

Therefore, 99.7% of the women over seventy have blood pressures between 104 mmHg and 158 mmHg.

• (b). Let's find the percentage of the women with blood pressures between 122 mmHg and 140 mmHg.

Using the z-score formula:

[tex]\begin{gathered} P(122

Thus, we have:

P(z ≤ 1) - P(z ≤ -1)

Using the standard normal distribution table, we have:

NORMSDIST(1) = 0.8413

NORMSDIST(-1) = 0.1587

P(z ≤ 1) - P(z ≤ -1) = 0.8413 - 0.1587 = 0.6826 ≈ 68%

Therefore, approximately 68% of the women have blood pressures between 122 mmHg and 140 mmHg.

ANSWER:

(a). 104 mmHg and 158 mmHg.

(b). 68%