Chapter 6.2, Problem 47E

(a)

To determine

Mean of randomly selected variable Y .

Answer to Problem 47E

Mean of Y is ${47.3}^{\text{o}}\text{F}$ .

Explanation of Solution

Given information:

T : temperature in degree Celsius

Y : temperature in degree Fahrenheit

For T :

Mean,

  ${\mu }_T={8.5}^{\text{o}}\text{C}$

Standard deviation,

  ${\sigma }_T={2.25}^{\text{o}}\text{C}$

Recall that

  $F=\left(\frac{9}{5}\right)\text{C}+\text{32}$

Calculations:

We know that

The degrees in Fahrenheit are the degrees in Celsius multiplied by $\frac{9}{5}$ and increased by 32.

  $Y\left(\text{F}\right)=\frac{9}{5}T\left(\text{C}\right)+\text{32}$

Then

Every data value in the distribution of T is multiplied by the same constant $\frac{9}{5}$ and increased by the same constant 32.

If every data value is added by the same constant, the center of the distribution is also increased by the same constant.

Also,

If every data value is multiplied by the same constant, the center of the distribution is also multiplied by the same constant.

We know that

The mean is the measure of the center.

Thus,

The mean is multiplied by $\frac{9}{5}$ and increased by 32.

  ${\mu }_Y=\frac{9}{5}{\mu }_T+32=\frac{9}{5}\left(8.5\right)+32=15.3+32={47.3}^{\text{o}}\text{F}$

(b)

To determine

Interpretation and calculation for standard deviation of Y .

Answer to Problem 47E

Standard deviation of Y ,

  ${\sigma }_Y={4.05}^{\text{o}}\text{F}$

Explanation of Solution

Given information:

T : temperature in degree Celsius

Y : temperature in degree Fahrenheit

For T :

Mean,

  ${\mu }_T={8.5}^{\text{o}}\text{C}$

Standard deviation,

  ${\sigma }_T={2.25}^{\text{o}}\text{C}$

Recall that

  $F=\left(\frac{9}{5}\right)\text{C}+\text{32}$

Calculations:

We know that

The degrees in Fahrenheit are the degrees in Celsius multiplied by $\frac{9}{5}$ and increased by 32.

  $Y\left(\text{F}\right)=\frac{9}{5}T\left(\text{C}\right)+\text{32}$

Then

Every data value in the distribution of T is multiplied by the same constant $\frac{9}{5}$ and increased by the same constant 32.

If every data value is added by the same constant, the spread of the distribution is unaffected.

Also,

If every data value is multiplied by the same constant, the spread of the distribution is also multiplied by the same constant.

We know that

The standard deviation is the measure of the spread.

Thus,

The standard deviation is multiplied by $\frac{9}{5}$ .

  ${\sigma }_Y=\frac{9}{5}{\sigma }_T=\frac{9}{5}\left(2.25\right)={4.05}^{\text{o}}\text{F}$

Thus,

The temperature in Fahrenheit of the cabin at midnight on a randomly selected night varies on average by ${4.05}^{\text{o}}\text{F}$ from the mean temperature of ${47.3}^{\text{o}}\text{F}$ .

(c)

To determine

Probability for the midnight temperature in the cabin is less than ${40}^{\text{o}}\text{F}$ .

Answer to Problem 47E

Probability for the midnight temperature less than ${40}^{\text{o}}\text{F}$ is 0.0359.

Explanation of Solution

Given information:

Midnight temperature,

  $x={40}^{\text{o}}\text{F}$

From Part (a),

  ${\mu }_Y={47.3}^{\text{o}}\text{F}$

From Part (b),

  ${\sigma }_Y={4.05}^{\text{o}}\text{F}$

Calculations:

Calculate the z − score,

  $z=\frac{x-\mu }{\sigma }=\frac{40-47.3}{4.05}\approx -1.80$

Use normal probability table in the appendix, to find the corresponding probability.

See the row that starts with -1.8 and the column that starts with .00 of the standard normal probability table for $P\left(z<-1.80\right)$ .

  $P\left(x<40\right)=P\left(z<-1.80\right)=0.0359$

Thus,

Probability for the midnight temperature in the cabin is less than ${40}^{\text{o}}\text{F}$ is 0.0359.