Question

In a process for manufacturing glassware, glass stems are sealed by heating them in a flame. The temperature of the flame can be adjusted to one of five different settings. Here is the probability distribution of the flame temperature $X$(in degrees Celsius) for a randomly chosen glass:

$\mu X=550°C\sigma X=5.7°C$

(a) The target temperature is$550°C$. What are the mean and standard deviation of the number of degrees off target, $D=X-550$?

(b) A manager asks for results in degrees Fahrenheit. The conversion of X into degrees Fahrenheit is given by \(y=\frac{9}{5} X+32\). What are the mean μY and the standard deviation σY of the temperature of the flame in the Fahrenheit scale?

Short answer

(a)Mean=${\mu }_X=550$

Standard deviation=${\sigma }_X=5.7$

(b)Mean= $100°\mathrm{F}$

Standard deviation=$10.26°\mathrm{F}$

Step-by-step solution

Part (a) Step 1: Given information

Given in the question that, In a process for manufacturing glassware, glass stems are sealed by heating them in a flame. The temperature of the flame can be adjusted to one of five different settings. Here is the probability distribution of the flame temperature $X$(in degrees Celsius) for a randomly chosen glass:

We need to find the mean and standard deviation of the number of degrees off target,$D=X-550$, if the target temperature is $550°C$.

Part (a) Step 2: Explanation

The probability distribution is:

Temperature	$540°$	$545°$	$550°$	$555°$	$560°$
Probability	$0.1$	$0.25$	$0.3$	$0.25$	$0.1$

The equation is:

$D=X-550$

Mean and standard deviation are:

${\mu }_X=550$

${\sigma }_X=5.7$

Part (a) Step 3: Mean and standard deviation

The mean and standard deviation can be calculated as:

${\mu }_{X-550}={\mu }_X-550$

$=550-550$

$=0$
${\sigma }_{X-550}={\sigma }_X$

$=5.7$

The mean and standard deviation are $0$and $5.7$.

Part (b) Step 1: Given information

In a process for manufacturing glassware, glass stems are sealed by heating them in a flame. The temperature of the flame can be adjusted to one of five different settings. Here is the probability distribution of the flame temperature $X$(in degrees Celsius) for a randomly chosen glass:

We need to find the mean $\mu Y$and the standard deviation $\sigma Y$of the temperature of the flame

Part (b) Step 2: Explanation

The mean and standard deviation can be calculated as:

${\mu }_{\frac{9}{5}X+32}=\frac{9}{5}{\mu }_X+32$

$=\frac{9}{5}\left(550\right)+32$

$={100}^{\circ }\mathrm{F}$

${\sigma }_{\frac{9}{5}}X+32=\frac{9}{5}{\sigma }_X$

$=\frac{9}{5}\left(5.7\right)$

$={10.26}^{\circ }\mathrm{F}$.