Last year’s freshman class at Big State University totaled 5,307 students. Of those, 1,264 received a merit scholarship to help offset tuition costs their freshman year (although the amount varied per student). The amount a student received was N($3,454, $490). If the cost of full tuition was $4,300 last year, what percentage of students who received a merit scholarship did not receive enough to cover full tuition? (Round your answer to the nearest whole percent.)

Last year the number of students at Big State University were 5307. Among these 1264 students received merit scholarship. The distribution of merit scholarship was Normal with following mean and standard deviation:

Mean = μ = $ 3454

Standard Deviation = σ = 490

Cost of Full Tuition is $4300. We have to find what percentage of students did not receive Full Tuition fee. Since, the distribution is given to be Normal, we will use the concept of z-distribution to find the required percentage.

We will convert x = 4300 to z-score and then find what percentage of z values are less than the calculated z-score. The formula to calculate the z-score is:

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

x = 4300 converted to z-score will be:

[tex]z=\frac{4300-3454}{490}=1.73[/tex]

P(X < 4300) is equivalent to P(z < 1.73). From the z-table we can find:

P(z < 1.73) = 0.9582 = 95.82%

Since,

P(X < 4300) = P(z < 1.73)

We can say that 95.82% of the students received an amount lesser than $4300. So from here we can conclude that:

Among the students who received merit scholarship, 96% did not receive enough to cover full tuition

codiepienagoya codiepienagoya · Answer 2 · 2021-12-07T13:14:14+01:00

Following are the calculation for the percentage of students:

Given:

[tex]\to \mu=\$ 3,454 \\\\ \to \sigma =\$ 490\\\\ \to x= 4300 \\\\[/tex]

To find:

percentage of students=?

Solution:

[tex]\to P(X<x)\\\\[/tex]

[tex]\to P(x< \$4300) = p(\frac{X-\mu}{\sigma} < \frac{4300 -3454}{490}) \\\\[/tex]

Therefore

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

[tex]\to P(z< \frac{846}{490})\\\\\to P(z< 1.73)\\\\[/tex] Using z table

[tex]= 0.9581 \\\\[/tex]

Therefore, the percentage is "[tex]95.81\% \approx 96\%[/tex]".

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