Question

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard deviation of $4.44. 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% or 75% or 84% or 89%) of the fares lie between 20.17 dollars and 37.13 dollars.
b. According to Chebyshev's theorem, at least 84% of the fares lie between ____ dollars and_____ dollars . (Round your answer to 2 decimal places.)
c. Suppose that the distribution is bell-shaped. According to the empirical rule, approximately 99.7% of the fares lie between ___ dollars and ___ dollars .
d. Suppose that the distribution is bell-shaped. According to the empirical rule, approximately____ (68% or 75% or 95% or 99.7%) of the fares lie between 20.17 dollars and 37.13 dollars.

Answer 1

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)  

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

  $= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$

In point c)

$\to 99.7 \% \\ lie \ between=28.75 \pm z(.03)\times \sigma \\\\ =28.75\pm 2.97 \times 4.44\\\\=(28.75\pm 13.1)\\\\=(15.65,41.85)$

In point d)

using standard normal variate

$x=20.17\\\\ z=-2 \\\\x=37.13\\\\ z=2\\\\\to P(20.17<x<37.13)=P(-2<z<2)=0.95$