Respuesta :
Answer:
a) 4.18% of students scored above a 91
b) 20.61% of students scored below a 63
c) A grade of 52.75 is the cutoff for being required to attend these sessions
d) The cutoff to get an A is a grade of 86.08.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 72, \sigma = 11[/tex]
(a) What percent of students scored above a 91?
This is 1 subtracted by the pvalue of Z when X = 91. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{91 - 72}{11}[/tex]
[tex]Z = 1.73[/tex]
[tex]Z = 1.73[/tex] has a pvalue of 0.95682
1 - 0.9582 = 0.0482
4.18% of students scored above a 91
(b) What percent of students scored below a 63?
This is the pvalue of Z when X = 63. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{63 - 72}{11}[/tex]
[tex]Z = -0.82[/tex]
[tex]Z = -0.82[/tex] has a pvalue of 0.2061.
20.61% of students scored below a 63
(c) If the lowest 4% of students will be required to attend peer tutoring sessions, what grade is the cutoff for being required to attend these sessions?
This is the value of X when Z has a pvalue of 0.04. So it is X when Z = -1.75.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.75 = \frac{X - 72}{11}[/tex]
[tex]X - 72 = -1.75*11[/tex]
[tex]X = 52.75[/tex]
A grade of 52.75 is the cutoff for being required to attend these sessions
(d) If the highest 10% of students will be given a grade of A, what is the cutoff to get an A?
This is the value of X when Z has a pvalue of 1-0.1 = 0.9. So it is X when Z = 1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 72}{11}[/tex]
[tex]X - 72 = 1.28*11[/tex]
[tex]X = 86.08[/tex]
The cutoff to get an A is a grade of 86.08.
