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To help students improve their reading, a school district decides to implement a reading program. It is to be administered to the bottom 15% of the students in the district, based on the scores on a reading achievement exam given in the child's dominant language. The reading-score for the pooled students in the district is approximately normally distributed with a mean of 122 points, and standard deviation of 18 points. a. Find the probability of students score above 140 points? b. What is the 15th percentile of the students eligible for the program?

Respuesta :

Answer:

a) [tex]P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140-122}{18})=P(Z>1)[/tex]

And we can find this probability using the complement rule and the normal standard table or excel and we got:

[tex]P(Z>1)=1-P(Z<1)=1-0.8413= 0.1587 [/tex]

b) [tex]z=-1.036<\frac{a-122}{18}[/tex]

And if we solve for a we got

[tex]a=122 -1.036*18=103.35[/tex]

So the value of height that separates the bottom 15% of data from the top 85% is 103.35.  

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

[tex]X \sim N(122,18)[/tex]  

Where [tex]\mu=122[/tex] and [tex]\sigma=18[/tex]

We are interested on this probability

[tex]P(X>140)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

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

[tex]P(X>140)=P(\frac{X-\mu}{\sigma}>\frac{140-\mu}{\sigma})=P(Z>\frac{140-122}{18})=P(Z>1)[/tex]

And we can find this probability using the complement rule and the normal standard table or excel and we got:

[tex]P(Z>1)=1-P(Z<1)=1-0.8413= 0.1587 [/tex]

Part b

For this part we want to find a value a, such that we satisfy this condition:

[tex]P(X<a)=0.15[/tex]   (a)

[tex]P(X>a)=0.85[/tex]   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.15 of the area on the left and 0.85 of the area on the right it's z=-1.036. On this case P(Z<-1.036)=0.15 and P(z>-1.036)=0.85

If we use condition (b) from previous we have this:

[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.15[/tex]  

[tex]P(z<\frac{a-\mu}{\sigma})=0.15[/tex]

But we know which value of z satisfy the previous equation so then we can do this:

[tex]z=-1.036<\frac{a-122}{18}[/tex]

And if we solve for a we got

[tex]a=122 -1.036*18=103.35[/tex]

So the value of height that separates the bottom 15% of data from the top 85% is 103.35.  

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