Question

$\mathrm{X}$A glass act In a process for manufacturing glassware, glass stems are sealed by heating them in a flame. Let X be the temperature (in degrees Celsius) for a randomly chosen glass. The mean and standard deviation of X are ${\mu }_X=550°\mathrm{C}$and ${\sigma }_X=5.7°\mathrm{C}$.

a. Is tempeгature a discrete ог continuous гandom variable? Explain уоuг answer.

b. The target temperature is $550°\mathrm{C}$. What are the mean and standard deviation of the number of degrees off target, $D=X-550$?

c. A manager asks for results in degrees Fahrenheit. The conversion of X into degrees Fahrenheit is given by $\mathrm{Y}=95\mathrm{X}+{32}^Y=\frac{9}{5}X+32$. What are the mean ${\mu }_Y$and the standard deviation ${\sigma }_Y$of the temperature of the flame in the Fahrenheit scale?

Short answer

(a)The temperature is a continuous random variable because it takes on decimal values i.e. $30.7°\mathrm{C}$

(b)Mean of D,

${\mu }_D={\mu }_{X-550}={\mu }_X-550=550-550=0°\mathrm{C}$

Standard deviation of D ,

${\sigma }_D={\sigma }_{X-550}={\sigma }_X=5.7°\mathrm{C}$

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

${\sigma }_Y=\frac{9}{5}{\sigma }_X=\frac{9}{5}(5.7)=10.26°\mathrm{F}$

Step-by-step solution

Part (a) Step 1: Given Information

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

For X :

Mean,

${\mu }_X=550°\mathrm{C}$

Standard deviation,

${\sigma }_X=5.7°\mathrm{C}$

Part (a) Step 2: Simplification

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. $30.7°\mathrm{C}$.

Part (b) Step 1: Given Information

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

For X:

Mean,

${\mu }_X=550°\mathrm{C}$

Standard deviation,

${\sigma }_X=5.7°\mathrm{C}$

Number of degrees off target,

$D=X-550$

Part (b) Step 2: Simplification

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,

${\mu }_D={\mu }_{X-550}={\mu }_X-550=550-550=0°\mathrm{C}$

Standard deviation of $D$,

${\sigma }_D={\sigma }_{X-550}={\sigma }_X=5.7°\mathrm{C}$

Part (c) Step 1: Given Information

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

For X:

Mean,

${\mu }_X=550°\mathrm{C}$

Standard deviation,

${\sigma }_X=5.7°\mathrm{C}$

Conversion of X nto degrees Fahrenheit:

$Y=\frac{9}{5}X+32$

Where,

Y: temperature in degrees Fahrenheit

Part (c) Step 2: Simplification

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 .

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

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 $\frac{9}{5}$.
${\sigma }_Y=\frac{9}{5}{\sigma }_X=\frac{9}{5}(5.7)=10.26°\mathrm{F}$