MATH SOLVE

4 months ago

Q:
# The gpa at the super university has a normal distribution with a mean 2.29 and standard deviation 1.29 to be an honors student at this university your GPA has to be in the top 20% what is the smallest gpa you need to have to be an honors student?

Accepted Solution

A:

The smallest GPA needed to fall in the top 20% is [tex]g[/tex], where

[tex]\mathbb P(G\ge g)=0.20[/tex]

where [tex]G[/tex] is a random variable denoting GPA scores. Transform [tex]G[/tex] to the standard normal random variable [tex]Z[/tex] using

[tex]Z=\dfrac{G-\mu_G}{\sigma_G}\iff G=\mu_G+\sigma_GZ[/tex]

where [tex]\mu_G[/tex] and [tex]\sigma_G[/tex] are the mean and standard deviation of [tex]G[/tex], respectively. Now

[tex]\mathbb P(G\ge g)=\mathbb P\left(Z\ge\dfrac{g-2.29}{1.29}\right)=0.20[/tex]

Using the inverse CDF for the standard normal distribution, we find that the corresponding critical value is about 0.8416, which means

[tex]\implies\dfrac{g-2.29}{1.29}\approx0.8416\implies g\approx3.3757[/tex]

[tex]\mathbb P(G\ge g)=0.20[/tex]

where [tex]G[/tex] is a random variable denoting GPA scores. Transform [tex]G[/tex] to the standard normal random variable [tex]Z[/tex] using

[tex]Z=\dfrac{G-\mu_G}{\sigma_G}\iff G=\mu_G+\sigma_GZ[/tex]

where [tex]\mu_G[/tex] and [tex]\sigma_G[/tex] are the mean and standard deviation of [tex]G[/tex], respectively. Now

[tex]\mathbb P(G\ge g)=\mathbb P\left(Z\ge\dfrac{g-2.29}{1.29}\right)=0.20[/tex]

Using the inverse CDF for the standard normal distribution, we find that the corresponding critical value is about 0.8416, which means

[tex]\implies\dfrac{g-2.29}{1.29}\approx0.8416\implies g\approx3.3757[/tex]