A major cab company in Chicago has computed its mean fare from O\'Hare Airport t

Question

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $25.80, with a standard deviation of $4.47. 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 between 19.095 dollars and 32.505 of the fares lie (b) According to Chebyshev's theorem, at least 2 between 16.86 dollars and 34.74 of the fares lie (c) Suppose that the distribution is bell-shaped. According to the empirical rule, approximately? 34.74 of the fares lie between 16.86 dollars and (d) Suppose that the distribution is bell-shaped. According to the empirical rule, approximately 68% of the fares lie between 0 dollars and dollars .

Explanation / Answer

Solution

Let X = fare ($) from O’Hare Airport to Drake Hotel.

Given mean = E(X) = 25.80 and Standard Deviation, = 4.47 ………………………………………….. (1)

Back-up Theory

Chebyshev’s Theorem

If E(X) = µ and V(X) = 2, then P(|X - µ| k) 1/k2 for a wide spectrum of distributions…………… (2)

For bell-shaped distribution, Empirical rule, also known as 68 – 95 – 99.7 percent rule:

P{(µ - ) X (µ + )} = 0.68;

P{ µ - 2) X (µ + 2)} = 0.95;

P{(µ - 3) X (µ + 3)} = 0.997 ................................................................…………………………….(3)

Now to work out the solution,

Preparatory Work

16.86 = µ - 2 ……………………………………………………………………………………………....….(4)

19.095 = µ - 1.5 ……………………………………………………………………………………………….(5)

32.505 = µ + 1.5 .………………………………………………………………………………………….….(6)

34.74 = µ + 2 …………………………………………………...…………………………………………….(7)

Part (a)

P(19.095 X 32.505)

= P[(µ - 1.5) X (µ + 1.5)], [vide (5) and (6)], which is same as

= P[- 1.5 (X - µ) 1.5], which is same as

= P[|(X - µ)| 1.5] …………………………………………………………………………………………….(8)

Comparing (2) and (8), k = 1.5. So, by Chebyshev’s Theorem

P[|(X - µ)| 1.5] {1/(1.5)2} or

P[|(X - µ)| 1.5] 0.4444

=> P[|(X - µ)| 1.5] 1 – 0.4444 = 0.5556

=> at least 55.56% of fares would lie between 19.095 and 32.505 ANSWER

Part (b)

P(16.86 X 34.74)

= P[(µ - 2) X (µ + 2)], [vide (4) and (7)], which is same as

= P[- 2 (X - µ) 2], which is same as

= P[|(X - µ)| 2] …………………………………………………………………………………………….(9)

Comparing (2) and (9), k = 2. So, by Chebyshev’s Theorem

P[|(X - µ)| 2] {1/(2)2} or

P[|(X - µ)| 2] 0.25

=> P[|(X - µ)| 2] 1 – 0.25 = 0.75

=> at least 75% of fares would lie between 16.86 and 34.74 ANSWER

Part (c)

P(16.86 X 34.74)

= P[(µ - 2) X (µ + 2)] [vide (4) and (7)],

= 0.95 [vide (3)]

=> 75% of fares would lie between 16.86 and 34.74 ANSWER

Part (d)

Vide (3), P{(µ - ) X (µ + )} = 0.68

=> 68% of fares would lie between (µ - ) and (µ + ) i.e., 21.33 and 30.27 ANSWER

DONE

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