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.

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

$$\begin{gathered} 渭=44and蟽=1.5 \\ x=43 \end{gathered}$$

The z-score is,

$\begin{array}{l}z=\frac{x-渭}{蟽}\\ =\frac{43-44}{1.5}\\ =-0.6667\end{array}$

$$\begin{gathered} P\left(x>43\right)=1-P\left(x<43\right) \\ =1-P\left(z<-0.6667\right) \\ =1-\left(1-P\left(z<-0.6667\right)\right) \\ =1-1+0.748511 \\ =0.748511 \\ 鈮�0.7485 \\ P\left(x>43\right)=0.7485 \end{gathered}$$

Thus, the required probability is 0.7485.

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

$$\begin{gathered} 渭=44and蟽=1.5 \\ x=42 \end{gathered}$$

The z-score is,

$\begin{array}{l}z=\frac{x-渭}{蟽}\\ =\frac{45-44}{1.5}\\ =-1.3333\end{array}$

Again,

$x=45$

The z-score is,

$\begin{array}{l}z=\frac{x-渭}{蟽}\\ =\frac{45-44}{1.5}\\ =0.6667\end{array}$

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

Thus, the required probability is 0.6567.

c.

No.

Explanation is given by,

If, the value belongs to$2蟽$limit, then it will be consideredas a usual value,

Then the interval of the$2蟽$limit is given by,

$$\begin{gathered} \left(渭-2蟽,渭+2蟽\right)=\left(\left(44-\left(2脳1.5\right)\right),\left(44-\left(2脳1.5\right)\right)\right) \\ =\left(\left(44-3\right),\left(44+3\right)\right) \\ =\left(41,47\right) \end{gathered}$$

Here, the observed response of 35 to an ecolabel does not belong to role="math" localid="1660286273313" $2蟽$limit.

So, ecolabel will not consider an Energy Star.