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

a)
`P(X>300)=P((X-\mu)/(\sigma)>(300-\mu)/(\sigma))=P(Z>(300-300)/(25))=P(z>0)= 0.5`

`P(X>335)=P((X-\mu)/(\sigma)>(335-\mu)/(\sigma))=P(Z>(335-300)/(25))=P(z>1.4)=0.0808`

b)
`P(\bar X>300)=P((\bar X-\mu)/(\sigma_(\bar x))>(300-\mu)/(\sigma_(\bar x)))=P(Z>(300-300)/(17.5))=P(z>0)= 0.5`

`P(\bar X>335)=P((\bar X-\mu)/(\sigma_(\bar x))>(335-\mu)/(\sigma_(\bar x)))=P(Z>(335-300)/(17.5))=P(z>2)=0.0228`

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=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X>300)=P((X-\mu)/(\sigma)>(300-\mu)/(\sigma))=P(Z>(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((X-\mu)/(\sigma)>(335-\mu)/(\sigma))=P(Z>(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) = (35)/(√(4))=17.5)`

The new z score is given by:

`z=(\bar X -\mu)/(\sigma_(\bar x))`

And using the formula we got:

`P(\bar X>300)=P((\bar X-\mu)/(\sigma_(\bar x))>(300-\mu)/(\sigma_(\bar x)))=P(Z>(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((\bar X-\mu)/(\sigma_(\bar x))>(335-\mu)/(\sigma_(\bar x)))=P(Z>(335-300)/(17.5))=P(z>2)=0.0228`