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The college board reported the following mean scores for the three parts of the scholastic aptitude test (sat) (the world almanac, 2009): critical reading 502 mathematics 515 writing 494assume that the population standard deviation on each part of the test is = 100.a. what is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the critical reading part of the test (to 4 decimals)? b. what is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the mathematics part of the test (to 4 decimals)? c. what is the probability a sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test (to 4 decimals)?

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Answer from: chrisandthemike76

a) P(492

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 0.949)= P(Z

b) P(505

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 1.898)= P(Z

c) P(484

And we can use excel or the normal standard table to find this probability:

P(-1 < Z< 1)= P(Z

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 scores for critical reading of a population, and for this case we know the distribution for X is given by:

$X \sim N(502,100)$

Where $\mu=502$ and $\sigma=100$

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)$

And for this case we want this probability:

$P(502-10 < \bar X < 502+10)$

And for this case we can use the z score given by:

$z= \frac{\bar X -\mu}{\sigma_{\bar x}}$

And if we use this formula we got:

P(492

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 0.949)= P(Z

Part b

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

$X \sim N(515,100)$

Where $\mu=515$ and $\sigma=100$

We select a sample of size n=90, since the distribution for X is normal then the distribution for the sample size is also normal

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{90}}=10.54)$

And for this case we want this probability:

$P(515-10 < \bar X < 515+10)$

And for this case we can use the z score given by:

$z= \frac{\bar X -\mu}{\sigma_{\bar x}}$

And if we use this formula we got:

P(505

And we can use excel or the normal standard table to find this probability:

P(-0.949 < Z< 1.898)= P(Z

Part c

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

$X \sim N(494,100)$

Where $\mu=494$ and $\sigma=100$

We select a sample of size n=100, since the distribution for X is normal then the distribution for the sample size is also normal

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}=\frac{100}{\sqrt{100}}=10)$

And for this case we want this probability:

$P(494-10 < \bar X < 494+10)$

And for this case we can use the z score given by:

$z= \frac{\bar X -\mu}{\sigma_{\bar x}}$

And if we use this formula we got:

P(484

And we can use excel or the normal standard table to find this probability:

P(-1 < Z< 1)= P(Z

Answer from: Quest

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Answer from: Quest

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