Question 461058
I'm assuming (and hoping) you have a calculator that can handle basic stats and deal with distributions. I'm going to assume you have a TI83 or a TI84.


a)


For this part, we're going to use the <a href="http://tibasicdev.wikidot.com/normalcdf">normalcdf</a> function. The function can be found by pressing the 2ND key and then the DISTR key to access the distribution menu. Then selection option 2 to select normalcdf(


After the parenthesis, enter the mean (in this case it's 510), then a comma followed by the standard deviation (in this case it's 80). Add another comma and then enter the boundaries. The boundaries in this case are 430 and 590. Separate each number by a comma. Finally, end everything by entering a )



So in the end, it should look like this


normalcdf(510,80,430,590)



Hit enter to compute and you'll find that the answer is approximately 0.68268949



This area represents the proportion (ie percentage) of people who scored in the range of 430 to 510. Multiply this value by the total number of people to get



0.68268949*50000 = 34,134.4745



and then round to the nearest whole number to get 34,134



So roughly 34,134 people scored between 430 and 590


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b)


Again, we're dealing with the normalcdf command (see above). So we must find the area under the normal distribution curve that has boundaries of 750 (on the left) and infinity (on the right). Since infinity isn't a number, we'll use 10000 in its place (it's big enough for our problem).



So type in 


normalcdf(510,80,750,10000)



to get the approximate answer of 0.001349898



Now convert to a percentage by multiplying by 100 to get 0.1349898%



So 0.1349898% of the students will qualify for a full scholarship.