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Scores on an exam follow an approximately normal distribution with a mean of 76. 4 and a standard deviation of 6. 1 points. What percent of students scored above 75 points?.

Respuesta :

Using the normal distribution, it is found that 59.09% of students scored above 75 points.

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 [tex]\mu = 76.4[/tex].
  • The standard deviation is [tex]\sigma = 6.1[/tex].

The proportion of students who scored above 75 points is 1 subtracted by the p-value of Z when X = 75, hence:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{75 - 76.4}{6.1}[/tex]

[tex]Z = -0.23[/tex]

[tex]Z = -0.23[/tex] has a p-value of 0.4091.

1 - 0.4091 = 0.5909

0.5909 = 59.09% of students scored above 75 points.

You can learn more about the normal distribution at https://brainly.com/question/24663213

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