Answer:
59%
Step-by-step explanation:
Mean = [tex]\mu = 500[/tex]
Standard deviation = [tex]\sigma = 110[/tex]
We are supposed to find 0 what is the probability that a randomly selected student has a test score between 350 and 550
Formula : [tex]z = \frac{x-\mu}{\sigma}[/tex]
at x = 350
[tex]z = \frac{350-500}{110}[/tex]
[tex]z =−1.36[/tex]
at x = 550
[tex]z = \frac{550-500}{110}[/tex]
[tex]z =0.454[/tex]
So, P(-1.36<z<0.454)
P(z<0.45)-P(z<-1.36)
Using z score table
= 0.6736-0.0869
= 0.5867
= 58.6% ≈ 59%
Thus the probability that a randomly selected student has a test score between 350 and 550 is 59%
