Answer :
Using the normal distribution, it is found that 4.18% of the students who took the SAT math test meet this requirement.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable that has mean given by [tex]\mu[/tex] and standard deviation represented by [tex]\sigma[/tex] is by the following equation:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is above(in case the score is positive) or below(in case the score is negative) the mean.
- From the z-score table, the p-value associated with the z-score is found, which represents the percentile of the measure X.
The mean and the standard deviation for the distribution of SAT scores is given as follows:
[tex]\mu = 528, \sigma = 117[/tex]
The proportion above 730 is given by one subtracted by the p-value of Z when X = 730, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
Z = (730 - 528)/117
Z = 1.73
Z = 1.73 has a p-value of 0.9582.
1 - 0.9582 = 0.0418.
Hence 4.18% of the students who took the SAT math test meet this requirement.
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Final answer:
The application of the statistics concept of normal distribution or z-score allows us to calculate that approximately 4.27% of the students meet the University of Michigan's recommended SAT Math score of at least 730.
Explanation:
The subject of the question is the application of Statistics in the Mathematics field, particularly regarding the calculation of the normal distribution (z-score) in a High School grade problem.
To calculate the percent of students who meet the University of Michigan's recommended SAT math score, we have to calculate the z-score for the given score (730). The z-score formula is:
Z = (X - μ) / σ
where X is the value we are comparing (in this case, the score 730), μ is the mean, and σ is the standard deviation. Applying the numbers:
Z = (730 - 528) / 117 ≈ 1.7265
The z-table or standard normal table will give us the area under the curve to the left of the z-score. For a z-score of 1.7265, the area (or the proportion of students scoring below 730) is approximately 0.9573 or 95.73%. Therefore, the percentage of students who meet or exceed the score of 730 is 100% - 95.73% = 4.27%.
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