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The set of life spans of an appliance is normally distributed with a mean mc012-1.jpg = 48 months and a standard deviation mc012-2.jpg = 8 months. What is the z-score of an appliance that stopped working at 64 months?

Respuesta :

InesWalston

Answer:

The z-score of an appliance that stopped working at 64 months is 2

Step-by-step explanation:

We know that,

[tex]Z=\dfrac{X-\mu}{\sigma}[/tex]

where,

Z = Z score,

X = raw score = 64

μ = mean = 48

σ = standard deviation = 8

Putting the values,

[tex]Z=\dfrac{64-48}{8}=\dfrac{16}{8}=2[/tex]

Therefore, the z-score of an appliance that stopped working at 64 months is 2.

MrRoyal

The z-score of an appliance that stopped working at 64 months is 2

The given parameters are:

Mean, [tex]\mu[/tex] = 48 months

Standard deviation, [tex]\sigma[/tex] = 8 months

The z-score is then calculated as:

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

For an appliance that stopped working at 64 months, we have:

x = 64

So, the equation becomes

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

Evaluate the differences

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

Evaluate the quotient

[tex]z = 2[/tex]

Hence, the z-score of an appliance that stopped working at 64 months is 2

Read more about normal distribution at:

https://brainly.com/question/4079902

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