Question 1197052: Last year's freshman class at Big State University totaled 5,327 students. Of those, 1,265 received a merit scholarship to help offset tuition costs their freshman year (although the amount varied per student). The amount received was N($3,468, $489). If the cost of full tuition was $4,100 last year, what percentage of students who received a merit scholarship did not received enough to cover full tuition? (Round your answer to the nearest whole percent.)
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
The notation N(3468, 489) tells us we have a normal distribution with these parameters
mu = 3468 = mean
sigma = 489 = standard deviation
If a student got a scholarship, then they landed somewhere on this normal distribution.
The majority of the students will cluster around the mean of $3468
Let's convert the raw score x = 4100 to its corresponding z score
z = (x - mu)/sigma
z = (4100 - 3468)/489
z = 1.29 approximately
Then we'll use a Z table such as this one
https://www.ztable.net/
or one found in the back of your book.
That table says
P(Z < 1.29) = 0.90147
So about 90.147% of the students who got a scholarship received less than $4,100
If you were to use a calculator such as this one
https://davidmlane.com/normal.html
then you'll find that P(Z < 1.29) = 0.9015
The area under the curve to the left of z = 1.29 is roughly 0.9015
Alternatively you can use a TI83 or TI84 calculator to type in normalcdf(-99,1.29) and it will display 0.9014746057 approximately. The normalcdf function can be found by hitting the key labeled "2nd" then pressing the VARS key.
If you wish to use a spreadsheet, then type in =NORM.DIST(1.29,0,1,true)
The 1.29 refers to the z score we calculated
the 0 and 1 are the mean and standard deviation respectively
The "true" tells the spreadsheet to compute the cdf rather than pdf
You could also type in =NORM.DIST(4100,3468,489,true) if you wanted to skip converting to a z score. This avoids the slight error that happens when we rounded the z score to 1.29
As you can see, there are various methods to calculating the area under the standard normal curve to the left of z = 1.29
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Answer: Approximately 90% (when rounding to the nearest whole percent)
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