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Sagot :
Answer:
a) The 36th percentile of the scores is of 74.68.
b) The 70th percentile of scores is 85.3.
c) The minimum score needed to get an A is 93.1.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score 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 p-value, we get the probability that the value of the measure is greater than X.
Mean of 79 and a standard deviation of 12.
This means that [tex]\mu = 79, \sigma = 12[/tex]
(a) Find the 36th percentile of the scores.
This is X when Z has a pvalue of 0.36. So X when Z = -0.36.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.36 = \frac{X - 79}{12}[/tex]
[tex]X - 79 = -0.36*12[/tex]
[tex]X = 74.68[/tex]
The 36th percentile of the scores is of 74.68.
(b) Find the 70th percentile of the scores.
This 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 - 79}{12}[/tex]
[tex]X - 79 = 0.525*12[/tex]
[tex]X = 85.3[/tex]
The 70th percentile of scores is 85.3.
(c) The instructor wants to give an A to the students whose scores were in the top 12% of the class. What is the minimum score needed to get an A?
The 100 - 12 = 88th percentile, which is X when Z has a pvalue of 0.88, so X when Z = 1.175.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.175 = \frac{X - 79}{12}[/tex]
[tex]X - 79 = 1.175*12[/tex]
[tex]X = 93.1[/tex]
The minimum score needed to get an A is 93.1.
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