Amanda earned a score of 940 on a national achievement test that was normally distributed. The mean test score was 850 with a standard deviation of 100. What proportion of students had a higher score than Amanda? Use your z table

Respuesta :

Using the normal distribution, it is found that 0.1841 = 18.41% of students had a higher score than Amanda.

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.

The mean and the standard deviation for the grades are given as follows:

[tex]\mu = 850, \sigma = 100[/tex]

The proportion of students that had a higher grade than Amanda is one subtracted by the p-value of Z when X = 940, hence:

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

[tex]Z = \frac{940 - 850}{100}[/tex]

Z = 0.9

Z = 0.9 has a p-value of 0.8159.

1 - 0.8159 = 0.1841.

0.1841 = 18.41% of students had a higher score than Amanda.

More can be learned about the normal distribution at https://brainly.com/question/15181104

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