Chapter 2.1, Problem 29E

(a)

To determine

Mean temperature reading in degrees Celsius.

Answer to Problem 29E

Mean temperature reading in degrees Celsius is ${25}^{\text{o}}\text{C}$ .

Explanation of Solution

Given information:

  ${}^{\text{o}}\text{C}=\frac{5}{9}\text{F}-\frac{160}{9}$

Mean, ${\mu }_{\text{F}}={77}^{\text{o}}\text{F}$

Standard deviation,

  ${\sigma }_{\text{F}}=3^{\text{o}}\text{F}$

Since

  ${}^{\text{o}}\text{C}=\frac{5}{9}\text{F}-\frac{160}{9}$ ,

In order to convert every data value from $\text{F}$ to ${}^{\text{o}}\text{C}$ , every data value needs to be multiplied by $\frac{5}{9}$ and decreased by $\frac{160}{9}$ .

Since mean represents the center of the distribution.

When every data value is multiplied by $\frac{5}{9}$ , then the center of the distribution is also multiplied by $\frac{5}{9}$ .

Similarly,

When every data value is decreased by $\frac{160}{9}$ , then the center of the distribution is also decreased by $\frac{160}{9}$ .

Then

The new mean in ${}^{\text{o}}\text{C}$ is the mean in $\text{F}$ multiplied by $\frac{5}{9}$ and decreased by $\frac{160}{9}$ .

New mean,

  ${\mu }_{\text{C}}=\frac{5}{9}{\mu }_{\text{F}}-\frac{160}{9}=\frac{5}{9}\left(77\right)-\frac{160}{9}=\frac{385}{9}-\frac{160}{9}=\frac{225}{9}={25}^{\text{o}}\text{C}$

Thus,

The mean temperature reading in degrees Celsius is ${25}^{\text{o}}\text{C}$ .

(b)

To determine

Standard deviation of the temperature readings in degrees Celsius.

Answer to Problem 29E

Standard deviation,

  ${\sigma }_{\text{C}}\approx {1.6667}^{\text{o}}\text{C}$

Explanation of Solution

Given information:

  ${}^{\text{o}}\text{C}=\frac{5}{9}\text{F}-\frac{160}{9}$

Mean, ${\mu }_{\text{F}}={77}^{\text{o}}\text{F}$

Standard deviation,

  ${\sigma }_{\text{F}}=3^{\text{o}}\text{F}$

Since

  ${}^{\text{o}}\text{C}=\frac{5}{9}\text{F}-\frac{160}{9}$ ,

In order to convert every data value from $\text{F}$ to ${}^{\text{o}}\text{C}$ , every data value needs to be multiplied by $\frac{5}{9}$ and decreased by $\frac{160}{9}$ .

Since the standard deviation is the measure of variability.

When every data value is multiplied by $\frac{5}{9}$ , then the variability of the distribution is also multiplied by $\frac{5}{9}$ .

And

When every data value is decreased by $\frac{160}{9}$ , then the variability of the distribution will not be affected.

Then

The new standard deviation in ${}^{\text{o}}\text{C}$ is the standard deviation in $\text{F}$ multiplied by $\frac{5}{9}$ but not decreased by $\frac{160}{9}$ .

New standard deviation,

  ${\sigma }_{\text{C}}=\frac{5}{9}{\sigma }_{\text{F}}=\frac{5}{9}\left(3\right)=\frac{15}{9}=\frac{5}{3}\approx {1.6667}^{\text{o}}\text{C}$

Thus,

The standard deviation of the temperature readings in degrees Celsius is ${1.6667}^{\text{o}}\text{C}$ .