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Sagot :
Using the normal distribution, it is found that:
A. There is a 0.0228 = 2.28% probability that a randomly selected score is greater than 81 points.
B. 81.85% of students scores are between 69 and 78.
C. The student scored in the 99.87th percentile.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- The z-score measures how many standard deviations the measure is above or below the mean.
- Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.
In this problem, the mean and the standard deviation are given, respectively, as:
[tex]\mu = 75, \sigma = 3[/tex].
Item a:
The probability is one subtracted by the p-value of Z when X = 81, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{81 - 75}{3}[/tex]
Z = 2.
Z = 2 has a p-value of 0.9772.
1 - 0.9772 = 0.0228.
0.0228 = 2.28% probability that a randomly selected score is greater than 81 points.
Item b:
The proportion is the p-value of Z when X = 78 subtracted by the p-value of Z when X = 69, hence:
X = 78:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{78 - 75}{3}[/tex]
Z = 1.
Z = 1 has a p-value of 0.8413.
X = 69:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{69 - 75}{3}[/tex]
Z = -2.
Z = -2 has a p-value of 0.0228.
0.8413 - 0.0228 = 0.8185.
81.85% of students scores are between 69 and 78.
Item c:
The percentile is the p-value of Z when X = 84, multiplied by 100, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{84- 75}{3}[/tex]
Z = 3.
Z = 3 has a p-value of 0.9987.
The student scored in the 99.87th percentile.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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