Question

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%

Answer 1

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

`\begin{gathered} z=(x-\mu)/(\sigma) \\ z=(71-78)/(4) \\ z=-(7)/(4) \\ z_1=-1.75 \\ \end{gathered}`

For a score of 85

`\begin{gathered} z_2=(85-78)/(4) \\ z_2=(7)/(4) \\ z_2=1.75 \end{gathered}`

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%