Respuesta :
Answer:
81.86% of students in the course have a GPA between 2.9 and 3.8.
Step-by-step explanation:
We are given that the mean GPA of students in a course at UC Davis is 3.2 with a standard deviation of 0.3.
Assuming that the data follows normal distribution.
Let X = GPA of students in a course at UC Davis
So, X ~ Normal([tex]\mu=3.2,\sigma^{2} =0.3^{2}[/tex])
The z score probability distribution for normal distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = population mean GPA = 3.2
[tex]\sigma[/tex] = standard deviation = 0.3
Now, the probability that the students in the course have a GPA between 2.9 and 3.8 is given by = P(2.9 < X < 3.8)
P(2.9 < X < 3.8) = P(X < 3.8) - P(X [tex]\leq[/tex] 2.9)
P(X < 3.8) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{3.8-3.2}{0.3}[/tex] ) = P(Z < 2) = 0.97725
P(X [tex]\leq[/tex] 2.9) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{2.9-3.2}{0.3}[/tex] ) = P(Z [tex]\leq[/tex] -1) = 1 - P(Z < 1)
= 1 - 0.84134 = 0.15866
The above probability is calculated by looking at the value of x = 2 and x = 1 in the z table which has an area of 0.97725 and 0.84134 respectively.
Therefore, P(2.9 < X < 3.8) = 0.97725 - 0.15866 = 0.8186
Hence, 81.86% of students in the course have a GPA between 2.9 and 3.8.