91影视

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.