Question

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)?

Answer 1

Answer:

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:

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