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A certain standardized test's math scores have a bell-shaped distribution with a mean of 525 and a standard deviation of 119. Complete parts (a) through (c)
(a) What percentage of standardized test scores is between 406 and 644?
(Round to one decimal place as needed.)
(b) What percentage of standardized test scores is less than 406 or greater than 644?
(Round to one decimal place as needed.)
(c) What percentage of standardized test scores is greater than 763?
Round to one decimal place as needed.)

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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