Question

A study was designed to investigate the effects of two​ variables, (1) A​ student's level of mathematical anxiety and​ (2) teaching​ method, on a​ student's achievement in a mathematics course.
Students who had a low level of mathematical anxiety were taught using the traditional expository method. These students obtained a mean score of 440 with a standard deviation of 20 on a standardized test.
1. Assuming a​ bell-shaped distribution, what percentage of scores exceeded 400​?

Answer 1

Answer:

$P(X>400)=P(\frac{X-\mu}{\sigma}>\frac{400-\mu}{\sigma})=P(Z>\frac{400-440}{20})=P(z>-2)$

And we can find this probability using the complement rule:

$P(z>-2)=1-P(z$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(z>-2)=1-P(z$

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

$X \sim N(440,20)$  

Where $\mu=440$ and $\sigma=20$

We are interested on this probability

$P(X>440)$

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:

$P(X>400)=P(\frac{X-\mu}{\sigma}>\frac{400-\mu}{\sigma})=P(Z>\frac{400-440}{20})=P(z>-2)$

And we can find this probability using the complement rule:

$P(z>-2)=1-P(z$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

$P(z>-2)=1-P(z$