Respuesta :
Answer:
The lowest score on the final exam that would qualify a student for an A is 80.
The lowest score on the final exam that would qualify a student for a B is 74.68.
The lowest score on the final exam that would qualify a student for a C is 67.33.
The lowest score on the final exam that would qualify a student for a D is 62.
Step-by-step explanation:
When the distribution is normal, we use 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.
Mean of 71 and a standard deviation of 7.
This means that [tex]\mu = 71, \sigma = 7[/tex]
Grades are assigned such that the top 10% receive A's, the next 20% received B's, the middle 40% receive C's, the next 20% receive D's, and the bottom 10% receive F's.
This means that:
90th percentile and above: A
70th percentile and below 90th: B
30th percentile to the 70th percentile: C
10th percentile to the 30th: D
Lowest score for an A:
Top 10% receive A, which means that the lowest score that would qualify a student for an A is the 100 - 10 = 90th percentile, which is X when Z has a pvalue of 0.9, so X when Z = 1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.28 = \frac{X - 71}{7}[/tex]
[tex]X - 71 = 7*1.28[/tex]
[tex]X = 80[/tex]
The lowest score on the final exam that would qualify a student for an A is 80.
Lowest score for a B:
70th percentile, which is X when Z has a pvalue of 0.7, so X when Z = 0.525.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.525 = \frac{X - 71}{7}[/tex]
[tex]X - 71 = 7*0.525[/tex]
[tex]X = 74.68[/tex]
The lowest score on the final exam that would qualify a student for a B is 74.68.
Lowest score for a C:
30th percentile, which is X when Z has a pvalue of 0.3, so X when Z = -0.525.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.525 = \frac{X - 71}{7}[/tex]
[tex]X - 71 = 7*(-0.525)[/tex]
[tex]X = 67.33[/tex]
The lowest score on the final exam that would qualify a student for a C is 67.33.
Lowest score for a D:
10th percentile, which is X when Z has a pvalue of 0.1, so X when Z = -1.28.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{X - 71}{7}[/tex]
[tex]X - 71 = 7*(-1.28)[/tex]
[tex]X = 62[/tex]
The lowest score on the final exam that would qualify a student for a D is 62.
