Hotels’ use of ecolabels. Refer to the Journal of Vacation Marketing (January 2016) study of travelers’ familiarity with ecolabels used by hotels, Exercise 2.64 (p. 104). Recall that adult travelers were shown a list of 6 different ecolabels, and asked, “Suppose the response is measured on a continuous scale from 10 (not familiar at all) to 50 (very familiar).” The mean and standard deviation for the Energy Star ecolabel are 44 and 1.5, respectively. Assume the distribution of the responses is approximately normally distributed.

a. Find the probability that a response to Energy Star exceeds 43.

b. Find the probability that a response to Energy Star falls between 42 and 45. c. If you observe a response of 35 to an ecolabel, do you think it is likely that the ecolabel was Energy Star? Explain.

Referring to exercise 2.23, the mean and standard deviation for the Energy Star ecolabel are 44 and 1.5 respectively.

Assume the x is approximately normally distributed.

a.

Here ,the mean and standard deviation of the random variable x is given by,

\(\mu = 44\,\,and\,\,\sigma = 1.5\)

\(x = 43\)

The z-score is,

\(\begin{aligned}z &= \frac{{x - \mu }}{\sigma }\\ &= \frac{{43 - 44}}{{1.5}}\\ &= - 0.6667\end{aligned}\)

\(\begin{aligned}P\left( {x > 43} \right) &= 1 - P\left( {x < 43} \right)\\ &= 1 - P\left( {z < - 0.6667} \right)\\ &= 1 - \left( {1 - P\left( {z < 0.6667} \right)} \right)\\ &= 1 - 1 + 0.7485711\\ &= 0.748511\\ \approx 0.7485\end{aligned}\)

\(P\left( {x > 43} \right) = 0.7485\)

TThus, the required probability is 0.7485.

b.

Here the mean and standard deviation of the random variable x is given by,

\(\mu = 44\,\,and\,\,\sigma = 1.5\)

\(x = 42\)

The z-score is,

\(\begin{aligned}z &= \frac{{x - \mu }}{\sigma }\\ &= \frac{{42 - 44}}{{1.5}}\\ &= - 1.3333\end{aligned}\)

Again,

\(x = 45\)

The z-score is,

\(\begin{aligned}z &= \frac{{x - \mu }}{\sigma }\\ &= \frac{{45 - 44}}{{1.5}}\\ &= 0.6667\end{aligned}\)

\(\begin{aligned}P\left( {42 < x < 45} \right) &= P\left( {x < 45} \right) - P\left( {x < 42} \right)\\ &= P\left( {z < 0.6667} \right) - P\left( {z < - 1.3333} \right)\\ &= P\left( {z < 0.6667} \right) - \left( {1 - P\left( {z < 1.3333} \right)} \right)\\ &= 0.7485 - \left( {1 - 0.9082} \right)\\ &= 0.7485 - 1 + 0.9082\\ &= 0.6567\end{aligned}\)

\(P\left( {42 < x < 45} \right) = 0.6567\)

Thus, the required probability is 0.6567.

c.

No.

Explanation is given by,

If, the value belongs to \(2\sigma \) limit, then it will be consideredas a usual value,

Then the interval of the \(2\sigma \) limit is given by,

\(\begin{aligned}\left( {\mu - 2\sigma ,\mu + 2\sigma } \right) &= \left( {\left( {44 - \left( {2 \times 1.5} \right)} \right),\left( {44 + \left( {2 \times 1.5} \right)} \right)} \right)\\ &= \left( {\left( {44 - 3} \right),\left( {44 + 3} \right)} \right)\\ &= \left( {41,47} \right)\end{aligned}\)

Here, the observed response of 35 to an ecolabel does not belong to \(2\sigma \) limit.

So, ecolabel will not consider an Energy Star.