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Scores on a quiz are normally distributed with a mean of 78 and a standard deviation of 4. Find the percentage of scores for a randomly selected student with a score between 71 and 85.a) 91.98%b) 4.01%c) 3.67%d) 95.99%

Sagot :

Answer:

Explanations:

Given the following parameters;

Mean = 78

Standard deviation = 4

If a randomly selected student with a score between 71 and 85, then we need the probability P(71

Determine the z-score for each score

For a score of 71

[tex]\begin{gathered} z=\frac{x-\mu}{\sigma} \\ z=\frac{71-78}{4} \\ z=-\frac{7}{4} \\ z_1=-1.75 \\ \end{gathered}[/tex]

For a score of 85

[tex]\begin{gathered} z_2=\frac{85-78}{4} \\ z_2=\frac{7}{4} \\ z_2=1.75 \end{gathered}[/tex]

Determine the probability for the z-score values P(-1.75 < z < 1.75) using the standard table.

[tex]\begin{gathered} P(-1.75Hence the percentage of scores for a randomly selected student with a score between 71 and 85 is 91.98%
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