Chapter 6, Problem R6.2RE

(a)

To determine

Whether temperature is a discrete or continuous random variable.

Answer to Problem R6.2RE

Temperature is a continuous random variable.

Explanation of Solution

Given information:

X : temperature (in degrees Celsius) for a randomly chosen glass

For X :

Mean,

  ${\mu }_X={550}^{\text{o}}\text{C}$

Standard deviation,

  ${\sigma }_X={5.7}^{\text{o}}\text{C}$

Calculations:

Discrete data are restricted to define separate values.

For example,

Integers or counts.

Whereas,

Continuous data are not restricted to define separate values

For example,

Decimal, rational or real numbers.

In this case,

The temperature is a continuous random variable because it takes on decimal values i.e. ${30.7}^{\text{o}}\text{C}$ .

(b)

To determine

Mean and standard deviation of the number of degrees off target.

Answer to Problem R6.2RE

Mean of the number of degrees off target,

  ${\mu }_D=0^{\text{o}}\text{C}$

Standard deviation of the number of degrees off target,

  ${\sigma }_D={5.7}^{\text{o}}\text{C}$

Explanation of Solution

Given information:

X : temperature (in degrees Celsius) for a randomly chosen glass

For X :

Mean,

  ${\mu }_X={550}^{\text{o}}\text{C}$

Standard deviation,

  ${\sigma }_X={5.7}^{\text{o}}\text{C}$

Number of degrees off target,

  $D=X-550$

Calculations:

Property mean:

  ${\mu }_{aX+b}=a{\mu }_X+b$

Property standard deviation:

  ${\sigma }_{aX+b}=\left|a\right|{\sigma }_X$

Now,

We have

  $D=X-550$

Thus,

Mean of D ,

  ${\mu }_D={\mu }_{X-550}={\mu }_X-550=550-550=0^{\text{o}}\text{C}$

Standard deviation of D ,

  ${\sigma }_D={\sigma }_{X-550}={\sigma }_X={5.7}^{\text{o}}\text{C}$

(c)

To determine

Mean and standard deviation of the temperature of the flame in the Fahrenheit scale.

Answer to Problem R6.2RE

Mean in Fahrenheit scale,

  ${\mu }_Y={1022}^{\text{o}}\text{F}$

Standard deviation in Fahrenheit scale,

  ${\sigma }_Y={10.26}^{\text{o}}\text{F}$

Explanation of Solution

Given information:

X : temperature (in degrees Celsius) for a randomly chosen glass

For X :

Mean,

  ${\mu }_X={550}^{\text{o}}\text{C}$

Standard deviation,

  ${\sigma }_X={5.7}^{\text{o}}\text{C}$

Conversion of X into degrees Fahrenheit:

  $Y=\frac{9}{5}X+32$

Where,

Y : temperature in degrees Fahrenheit

Calculations:

For temperature conversion (degree Celsius to degree Fahrenheit):

  $Y=\frac{9}{5}X+32$

Then

Every data value in the distribution of Y 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.

  ${\mu }_Y=\frac{9}{5}{\mu }_X+32=\frac{9}{5}\left(550\right)+32=990+32={1022}^{\text{o}}\text{F}$

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 $\frac{9}{5}$ .

  ${\sigma }_Y=\frac{9}{5}{\sigma }_X=\frac{9}{5}\left(5.7\right)={10.26}^{\text{o}}\text{F}$