The set of life spans of an appliance is normally distributed with a mean [tex]\mu = 48[/tex] months and a standard deviation [tex]\sigma = 8[/tex] months.

What is the [tex]z[/tex]-score of an appliance that stopped working at 64 months?

A. [tex]-2[/tex]
B. [tex]-1[/tex]
C. [tex]1[/tex]
D. [tex]2[/tex]



Answer :

To find the [tex]$z$[/tex]-score of an appliance that stopped working at 64 months, we need to use the [tex]$z$[/tex]-score formula. The [tex]$z$[/tex]-score formula is:

[tex]\[ z = \frac{x - \mu}{\sigma} \][/tex]

where:
- [tex]\( x \)[/tex] is the value we are interested in (the lifespan of the appliance that stopped working),
- [tex]\( \mu \)[/tex] is the mean of the distribution,
- [tex]\( \sigma \)[/tex] is the standard deviation of the distribution.

Given:
- The mean [tex]\(\mu\)[/tex] is 48 months,
- The standard deviation [tex]\(\sigma\)[/tex] is 8 months,
- The value [tex]\( x \)[/tex] is 64 months.

We can plug these values into the formula:

[tex]\[ z = \frac{64 - 48}{8} \][/tex]

First, compute the numerator:

[tex]\[ 64 - 48 = 16 \][/tex]

Then, divide by the standard deviation:

[tex]\[ z = \frac{16}{8} = 2 \][/tex]

Therefore, the [tex]$z$[/tex]-score of an appliance that stopped working at 64 months is:

[tex]\[ 2 \][/tex]

So the correct answer is:

[tex]\[ \boxed{2} \][/tex]