Ann, who turns eighty years old this year, has just learned about blood pressure problems in the elderly and is interested in how her blood pressure compares to those of her peers. She has uncovered an article in a scientific journal that reports that the mean systolic blood pressure measurement for women over seventy-five is 130.1 mmHg, with a standard deviation of 6.8 mmHg.Assume that the article reported correct information. Complete the following statements about the distribution of systolic blood pressure measurements for women over seventy-five.(a) According to Chebyshev's theorem, at least $ \frac{8}{9} $ (about 89%) of the measurements lie between ⬜ mmHg and ⬜ mmHg. (Round your answer to 1 decimal place.)(b) According to Chebyshev's theorem, at least (Choose one) ▼ of the measurements lie between 116.5 mmHg and 143.7 mmHg.There is a box with options to fill in and buttons to reset or confirm selections next to the questions.

a) Suppose a random variable X defines the systolic blood pressure measurements of a woman over seventy five.In this case, the mean of X is μ = 130.1 mmHg and the standard deviation is σ = 6.8 mmHg. The formula for Chebyshev’s Inequality is: P(| X – μ | < kσ) ≥ 1 – (1/k2), where k is any positive real number. At least 89% of the data implies that P (| X – μ | < kσ) ≥ 8/9 (= (1 – (1/k2)). Hence, 1 – (1/k2) = 8/9 k2 = 1/(1 – 8/9) k2 = 9 k ≈ 3 Hence, at least 89% of the scores lie between 109.7 (= 130.1 – (3 × 6.8)) and 150.5 (= 130.1 + (3 × 6.8)). Thus, according to Chebyshev's theorem, at least 89% of the measurements lie between 109.7 mmHg and 150.5 mmHg. b) According to Chebyshev's theorem, the proportion of the measurements lie between 116.5 mmHg and 143.7 mmHg is, P(116.5 < X < 143.7) = P((116.5 – 130.1)/6.8 < (X – μ)/6.8 < (143.7 – 130.1)/6.8) = P(–2 < (X – μ)/6.8 < 2) = P (| X – μ | < (2)(6.8)) ≥ 1 – (1/22) [Here, k = 2] = 3/4 = 0.75. Hence, according to Chebyshev's theorem, the at least 75% (= 100 × 0.75) of the measurements lie between 116.5 mmHg and 143.7 mmHg.