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san4es73 [151]
3 years ago
12

Mark has a jar of marbles. The jar contains 1 red marble, 2 green marbles, 15 yellow marbles, and 11 blue marbles. Without looki

ng, Mark chooses a marble from the jar. Which statement is true?
Mathematics
2 answers:
Sphinxa [80]3 years ago
6 0

Answer:

24

Step-by-step explanation:

Vitek1552 [10]3 years ago
4 0

Answer:what are that statements

Step-by-step explanation:

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Dear math why don't you grow up and solve your own problems
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LOL

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Aiden has his two friends order a pizza with 12 slices together they ate 2/3 of the pizza if they evenly divided what was left h
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(03.02 HC) Stan, Liam, and Louise are competing in a cooking competition. They all used different amounts of flour from a can co
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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

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3 years ago
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Step-by-step explanation:

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