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Answer :
To find the maximum SAT score in 2014 that meets the course requirements for students in the lower 25% of scores, we need to determine the 25th percentile of the SAT score distribution.
The SAT scores in 2014 were approximately normally distributed with a mean (average) score, [tex]\(\mu = 1497\)[/tex], and a standard deviation, [tex]\(\sigma = 322\)[/tex].
In a normal distribution, the percentile indicates the value below which a given percentage of observations fall. Here, we want to find the score below which 25% of students fall, which is known as the 25th percentile.
To find the 25th percentile of a normal distribution, we use the properties of the normal distribution. Specifically, we look for the z-score that corresponds to the 25th percentile, which is typically around -0.674 (since 25% of the data is below this z-score).
1. Start by calculating the z-score for the 25th percentile with respect to the standard normal distribution, which is approximately -0.674.
2. Use the formula for converting a z-score to a raw score in a normal distribution:
[tex]\[
\text{SAT score} = \mu + z \times \sigma
\][/tex]
where [tex]\(\mu\)[/tex] is the mean, [tex]\(z\)[/tex] is the z-score, and [tex]\(\sigma\)[/tex] is the standard deviation.
3. Substitute the values:
- Mean, [tex]\(\mu = 1497\)[/tex]
- Standard deviation, [tex]\(\sigma = 322\)[/tex]
- Z-score for 25th percentile, [tex]\(z = -0.674\)[/tex]
[tex]\[
\text{SAT score} = 1497 + (-0.674 \times 322)
\][/tex]
4. Calculate:
[tex]\[
\text{SAT score} = 1497 - 217.1868 \approx 1279.81
\][/tex]
Therefore, the maximum SAT score in 2014 that qualifies for the course requirements is approximately 1279. This corresponds to the 25th percentile of the SAT score distribution for that year.
The SAT scores in 2014 were approximately normally distributed with a mean (average) score, [tex]\(\mu = 1497\)[/tex], and a standard deviation, [tex]\(\sigma = 322\)[/tex].
In a normal distribution, the percentile indicates the value below which a given percentage of observations fall. Here, we want to find the score below which 25% of students fall, which is known as the 25th percentile.
To find the 25th percentile of a normal distribution, we use the properties of the normal distribution. Specifically, we look for the z-score that corresponds to the 25th percentile, which is typically around -0.674 (since 25% of the data is below this z-score).
1. Start by calculating the z-score for the 25th percentile with respect to the standard normal distribution, which is approximately -0.674.
2. Use the formula for converting a z-score to a raw score in a normal distribution:
[tex]\[
\text{SAT score} = \mu + z \times \sigma
\][/tex]
where [tex]\(\mu\)[/tex] is the mean, [tex]\(z\)[/tex] is the z-score, and [tex]\(\sigma\)[/tex] is the standard deviation.
3. Substitute the values:
- Mean, [tex]\(\mu = 1497\)[/tex]
- Standard deviation, [tex]\(\sigma = 322\)[/tex]
- Z-score for 25th percentile, [tex]\(z = -0.674\)[/tex]
[tex]\[
\text{SAT score} = 1497 + (-0.674 \times 322)
\][/tex]
4. Calculate:
[tex]\[
\text{SAT score} = 1497 - 217.1868 \approx 1279.81
\][/tex]
Therefore, the maximum SAT score in 2014 that qualifies for the course requirements is approximately 1279. This corresponds to the 25th percentile of the SAT score distribution for that year.
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