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earnstyle [38]
3 years ago
8

Please help!!!!!!!!!!!!!!!!!!!!!!

Mathematics
1 answer:
Diano4ka-milaya [45]3 years ago
5 0
The right answer is B.
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I have Warnnnnnnnnn <br>Known<br>2, 8, 14, 20<br><br>Determine the term - 1000 ?​
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<h2>{\green {\boxed {\colorbox {black} {\green {King\:Brainly01}}}}} </h2>

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the recycling club collected 250 pounds of recyclables last semester. among the items collected, one fifths of weight could be c
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3 years ago
It is known that IQ scores form a normal distribution with a mean of 100 and a standard deviation of 15. If a researcher obtains
sdas [7]

Answer:

X \sim N(100,15)  

Where \mu=100 and \sigma=15

We select a sample of n=16 and we are interested on the distribution of \bar X, since the distribution for X is normal then we can conclude that the distribution for \bar X is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Because by definition:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}

And for this case we have this:

\mu_{\bar X}= \mu = 100

\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75

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".  

Solution to the problem

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

X \sim N(100,15)  

Where \mu=100 and \sigma=15

We select a sample of n=16 and we are interested on the distribution of \bar X, since the distribution for X is normal then we can conclude that the distribution for \bar X is also normal and given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Because by definition:

\bar X = \frac{\sum_{i=1}^n X_i}{n}

E(\bar X) = \mu

Var(\bar X) = \frac{\sigma^2}{n}

And for this case we have this:

\mu_{\bar X}= \mu = 100

\sigma_{\bar X} = \frac{15}{\sqrt{16}}= 3.75

6 0
3 years ago
You receive 9 text messages in 12 minuets what is the unit rate of text messages per hour
Dvinal [7]
9/12 which simplifies to 3/4 
give thanks and hope this helps
7 0
4 years ago
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