One of the most common examples of a normal distribution is the distribution of

Question

One of the most common examples of a normal distribution is the distribution of scores on standardized tests like the ACT. In 2010, the mean score was 21 and the standard deviation was 5.2 (Source: National Center for the Education Statistics). Use the information, along with the empirical rule and the approximate area under the curve, to sketch a normal distribution curve for this test. 1. What percentage of students scored below 21? 50% 2. About what percentage of students scored below 16? About 16%. About what percentage of students scored between 11 and 26? 13.5% + 34% + 34% + = 81.5% 4. Pulling all of the data for all test takers is a time-consuming and expensive practice. If we draw a sample instead, what do you think the mean would be? How confident are you in that value? 5. Your friend, Calvin, would like to go to a very selective college that only admits the top 1% of all student applicants. Calvin has good grades and scored 33 on the test. Do you think that Calvin's ACT score gives him a good chance of being admitted? Explain your answer. Z^* = 33 - 21/5.2 = z.30769 = 3 yes, because the normal cdf is 1.054%

Explanation / Answer

Solution :

Using the empirical rule , here 68% data lie between the (15.8,26.2) that is 1SD from mean. 95% data lie between the (10.6,31.4) that is 2SD from mean. And 99.7% data lie between the (5.4,36.6) that is 3SD from mean.

So please change the limits on right side of above normal curve and they are 26.2,31.4 and 36.6.

1) Percent of data below 21 = 50% ( as the area under total curve is 1, 50% data lie on left side and 50% data lie on right side from mean).

2) Percentage of data lie below 16 = P( x < 16) = P( z < -0.96)   

Using z-score formula = (x-mean)/standard deviation

So,Percentage of data lie below 16 = P( x < 16) = P( z < -0.96)   = 0.1685    (uisng the standard normal table)

3) Percent of data between 11 and 26 = P( 11 < x < 26) = P(-1.92 < z < 0.96)

Uisng the z-score formula we get above z-scores as -1.92 and 0.96

And now using rule P( a < z < b) = P( z < b) - P( z < a)

P(11 <x < 26) = P( -1.92 < z < 0.96) = P( z < 0.96) - P( z < -1.92) = 0.8315 - 0.0274 = 0.8041

So percent of data between 11 and 26 is 0.8041

4) Using the central limit theorem , as sample size increases sample data follows normla distribution with same mean.So here mean will remain same that is here Mean = 21.

5) Here we will find P( x < 33) = P( z < 2.3077) = P(z < 2.31) = 0.9896 (Uisng the table we get probability as 0.9896)

Yes , that is here Calvin's ACT score has 98.96% of chance being admitted.

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