Question

A certain standardized test's math scores have a bell-shaped distribution with a mean of 530 and a standard deviation of 119. Complete parts (a) through (c) a. What percentage of standardized test scores is between 411 and 649> % (Round to the nearest tenth as needed.) b. What percentage of standardized test scores is less than 411 or greater than 649? % (Round to the nearest tenth as needed.)c. What percentage of standardized test scores is greater dun 768? % (Round to the neatest tenth as needed.)

Answer 1

Answer:

a) $P(411$

And we can find this probability with this difference:

$P(-1$

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

$P(-1$

And the percentage would be 68.3%

b) For this case we can find this with the complement rule and with the result of the part a and we got:

$P(X649) = 1-P(411< X< 649) = 1-0.683 = 0.317$

And that represent 31.7%

c) $P(X>768)=P(\frac{X-\mu}{\sigma}>\frac{768-\mu}{\sigma})=P(Z>\frac{768-530}{119})=P(z>2)$

And we can find this probability with the complment rule like this:

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

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

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

And the percentage would be 2.3%

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".  

Part a

Let X the random variable that represent the standardized test's

Where $\mu=530$ and $\sigma=119$

We are interested on this probability

$P(411$

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:

$P(411$

And we can find this probability with this difference:

$P(-1$

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

$P(-1$

And the percentage would be 68.3%

Part b

For this case we can find this with the complement rule and with the result of the part a and we got:

$P(X649) = 1-P(411< X< 649) = 1-0.683 = 0.317$

And that represent 31.7%

Part c

$P(X>768)$

Using the z score we got:

$P(X>768)=P(\frac{X-\mu}{\sigma}>\frac{768-\mu}{\sigma})=P(Z>\frac{768-530}{119})=P(z>2)$

And we can find this probability with the complment rule like this:

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

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

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

And the percentage would be 2.3%