Y+2=-3X-4 in standard form
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Using the normal distribution, it is found that 62.46% of students would be expected to score between 350 and 550.
<h3>Normal Probability Distribution</h3>In a normal distribution with mean and standard deviation
, the z-score of a measure X is given by:
- 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:
The proportion of students that would be expected to score between 350 and 550 is the <u>p-value of Z when X = 550 subtracted by the p-value of Z when X = 350</u>, hence:
X = 550:
has a p-value of 0.8729.
X = 350:
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 brainly.com/question/24663213