Respuesta :
Using the normal distribution, it is found that 62.46% of students would be expected to score between 350 and 550.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
In this problem, the mean and the standard deviation are respectively, given by: [tex]\mu = 425, \sigma = 110[/tex]
The proportion of students that would be expected to score between 350 and 550 is the p-value of Z when X = 550 subtracted by the p-value of Z when X = 350, hence:
X = 550:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{550 - 425}{110}[/tex]
[tex]Z = 1.14[/tex]
[tex]Z = 1.14[/tex] has a p-value of 0.8729.
X = 350:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{350 - 425}{110}[/tex]
[tex]Z = -0.68[/tex]
[tex]Z = -0.68[/tex] has a p-value of 0.2483.
0.8729 - 0.2483 = 0.6246
62.46% of students would be expected to score between 350 and 550.
More can be learned about the normal distribution at https://brainly.com/question/24663213
