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3 years ago

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

Mathematics

1 answer:

ki77a [65]3 years ago

6 0

Answer:

$z=\frac{4415-3400}{450}=2.26$

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:

$X \sim N(3400,450)$

Where $\mu=3400$ and $\sigma=450$

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

$z=\frac{x-\mu}{\sigma}$

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

$z=\frac{4415-3400}{450}=2.26$

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

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Wires manufactured for use in a computer system are specified to have resistances between 0.11 and 0.13 ohms. The actual measure

Marina CMI [18]

Answer:

a) $P(0.11$

And we can find this probability with this difference and with the normal standard table or excel:

$P(-1.11$

b) $P(0.11 < \bar X < 0.13)$

And we can use the z score defined by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using the limits we got:

$z = \frac{0.11-0.12}{\frac{0.009}{\sqrt{4}}}= -2.22$

$z = \frac{0.13-0.12}{\frac{0.009}{\sqrt{4}}}= 2.22$

And we want to find this probability:

$P(-2.22< Z$

Step-by-step explanation:

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".

Part a

Let X the random variable that represent the resitances of a population, and for this case we know the distribution for X is given by:

$X \sim N(0.12,0.009)$

Where $\mu=0.12$ and $\sigma=0.009$

We are interested on this probability

$P(0.11$