Answer:
24.2% students received grades with z-scores between 0.15 and 0.85
Step-by-step explanation:
We are given the following in the question:
The grades of a benchmark test for North High School were normally distributed.
WE have to find the percentage of students that received grades with z-scores between 0.15 and 0.85.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(score between 0.15 < z < 0.85)
[tex]P(0.15 \leq z \leq 0.85)\\\\= P(z \leq 0.85) - P(z \leq 0.15)\\\\\text{Calculating the value from standard normal z-table}\\\\= 0.802 - 0.560 = 0.242 = 24.2\%[/tex]
24.2% students received grades with z-scores between 0.15 and 0.85
