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Sagot :
Using the normal distribution, it is found that:
a) 68.2% of standardized test scores are between 406 and 644.
b) 31.8% of standardized test scores are less than 406 or greater than 644.
c) 2.3% of standardized test scores are greater than 763.
Normal Probability Distribution
In a normal distribution 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]
- It 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, which is the percentile of X.
In this problem:
- The mean is of 525, hence [tex]\mu = 525[/tex].
- The standard deviation is of 119, hence [tex]\sigma = 119[/tex].
Item a:
The proportion is the p-value of Z when X = 644 subtracted by the p-value of Z when X = 406, hence:
X = 644:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{644 - 525}{119}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a p-value of 0.841.
X = 406:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{406 - 525}{119}[/tex]
[tex]Z = -1[/tex]
[tex]Z = -1[/tex] has a p-value of 0.159.
0.841 - 0.159 = 0.682.
0.682 = 68.2% of standardized test scores are between 406 and 644.
Item b:
Complementary event to the one found in item A, hence:
1 - 0.682 = 0.318.
0.318 = 31.8% of standardized test scores are less than 406 or greater than 644.
Item c:
The proportion is 1 subtracted by the p-value of Z when X = 763, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{763 - 525}{119}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] has a p-value of 0.977.
1 - 0.977 = 0.023
0.023 = 2.3% of standardized test scores are greater than 763.
You can learn more about the normal distribution at https://brainly.com/question/24663213
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