The number of text messsages that teenagers send per month is normally

distributed with a mean of 3,400 and a standard deviation of 450. If Kendra

sent 4,415 text messages last month, find her r-score to the nearest

hundredth.

Respuesta :

Answer:

[tex]z=\frac{4415-3400}{450}=2.26[/tex]

And the explanation of this number is:"The number of text messages for Kendra it's 2.26 deviations above the mean"

Step-by-step explanation:

1) Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

2) Calculate the z score

Let X the random variable that represent the number of text messages per month, and for this case we know the distribution for X is given by:

[tex]X \sim N(3400,450)[/tex]  

Where [tex]\mu=3400[/tex] and [tex]\sigma=450[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

If we apply this formula to our probability we got this:

[tex]z=\frac{4415-3400}{450}=2.26[/tex]

And the explanation of this number is:"The number of text messages for Kendra it's 2.26 deviations above the mean"