Question

The scores of 12th-grade students on the national assessment of educational progress year 2000 mathematics test have a distribution that is approximately normal with mean of 300 and standard deviation of 35.

Answer 1

Answer:

a)$P(X>300)=P(\frac{X-\mu}{\sigma}>\frac{300-\mu}{\sigma})=P(Z>\frac{300-300}{25})=P(z>0)= 0.5$

$P(X>335)=P(\frac{X-\mu}{\sigma}>\frac{335-\mu}{\sigma})=P(Z>\frac{335-300}{25})=P(z>1.4)=0.0808$

b)$P(\bar X>300)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}>\frac{300-\mu}{\sigma_{\bar x}})=P(Z>\frac{300-300}{17.5})=P(z>0)= 0.5$

$P(\bar X>335)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}>\frac{335-\mu}{\sigma_{\bar x}})=P(Z>\frac{335-300}{17.5})=P(z>2)=0.0228$

Step-by-step explanation:

Assuming the following questions:

a) Choose one twelfth-grader at random. What is the probability that his or her score is higher than 300? Higher than 335?

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".  

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

$X \sim N(300,35)$  

Where $\mu=300$ and $\sigma=35$

We are interested on this probability

$P(X>300)$

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:

$P(X>300)=P(\frac{X-\mu}{\sigma}>\frac{300-\mu}{\sigma})=P(Z>\frac{300-300}{25})=P(z>0)= 0.5$

We find the probabilities with the normal standard table or with excel.

And for the other case:

$P(X>335)=P(\frac{X-\mu}{\sigma}>\frac{335-\mu}{\sigma})=P(Z>\frac{335-300}{25})=P(z>1.4)=0.0808$

b) Now choose an SRS of four twelfth-graders. What is the probability that his or her mean score is higher than 300? Higher than 335?

For this case since the distribution for X is normal then the distribution for the sample mean is also normal and given by:

$\bar X = \sim N(\mu = 300, \sigma_{\bar x} = \frac{35}{\sqrt{4}}=17.5)$

The new z score is given by:

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

And using the formula we got:

$P(\bar X>300)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}>\frac{300-\mu}{\sigma_{\bar x}})=P(Z>\frac{300-300}{17.5})=P(z>0)= 0.5$

We find the probabilities with the normal standard table or with excel.

And for the other case:

$P(\bar X>335)=P(\frac{\bar X-\mu}{\sigma_{\bar x}}>\frac{335-\mu}{\sigma_{\bar x}})=P(Z>\frac{335-300}{17.5})=P(z>2)=0.0228$