Please help, struggling with question A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $27.90 with a standard deviation of $4.23. Based on this information, complete the following statements about the distribution of the company's fares from O'Hare Airport to the Drake Hotel.(a) According to Chebyshev's theorem, at least 56% of the fares lie between [ ] dollars and [ ] dollars. (Round your answer to 2 decimal places.)(b) According to Chebyshev's theorem, at least [Choose one] % of the fares lie between 19.44 dollars and 36.36 dollars. ---There are no graphs or diagrams included in the image. The content focuses on applying Chebyshev's theorem to determine the range within which a certain percentage of fares lie, pertaining to the cab fares in Chicago.

Consider that a random variable X has expected value μ and standard deviation, σ. Now, according to Chebyshev’s Inequality, it can be said that: P(| X – μ | < kσ) ≥ 1 – (1/k2), where k is any positive real number.a) Consider that the random variable X defines the fare from airport to hotel. In this case, the mean of X is μ = $27.90 and the standard deviation is σ = $4.23 At least 56% of the data implies that P (| X – μ | < kσ) ≥ 0.56 ( = (1 – (1/k2)). Hence, 1 – (1/k2) = 0.56 k2 = 1/(1 – 0.56) k2 = 2.27273 k = 1.5076 Hence, according to Chebyshev’s theorem, at least 56% of the observations lie between 21.52 (= 27.90 – (1.5076 × 4.23)) and 34.28 (= 27.90 + (1.5076 × 4.23)). Hence, according to Chebyshev’s theorem, at least 56% of the fares lie between 21.52 dollars and 34.28 dollars. b) By Chebyshev’s Inequality, P(19.44 < X < 36.36) = P((19.44 – 27.90)/4.23 < (X – μ)/4.23 < (36.36 – 27.90)/4.23) = P(–2 < (X – μ)/4.23 < 2) = P (| X – μ | < (2)(4.23)) ≥ 1 – (1/22) [Here, k = 2] = 3/4 = 0.75. Hence, according to Chebyshev’s theorem at least 75% (= 100 × 0.75) of the fares lie between 19.44 dollars and 36.36 dollars.