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Answer :
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The question asks us to find the percentage of students from a high school who score below 1148 on their SATs, knowing that the scores are normally distributed with a mean of 96 and a standard deviation of 290.
To solve this, follow these steps:
1. Understand the Normal Distribution and Z-Score:
- The normal distribution is symmetrically centered around the mean.
- The Z-score tells us how many standard deviations an element is from the mean.
2. Calculate the Z-Score:
- The formula for the Z-score is:
[tex]\[
Z = \frac{X - \mu}{\sigma}
\][/tex]
where [tex]\(X\)[/tex] is the score we are interested in (1148), [tex]\(\mu\)[/tex] is the mean (96), and [tex]\(\sigma\)[/tex] is the standard deviation (290).
3. Substitute the values into the Z-score formula:
- [tex]\[
Z = \frac{1148 - 96}{290} \approx 3.63
\][/tex]
4. Use the Z-Score to Find the Cumulative Probability:
- The cumulative probability associated with a Z-score of 3.63 is approximately 0.9999 (looking this up in standard normal distribution tables or using a calculator).
5. Calculate the Remaining Percentage:
- Since 0.9999 (or 99.99%) of the students score below 1148, we need to subtract this from 1 to find the percentage that meet this requirement.
- [tex]\[
\text{Percentage of students scoring below 1148} = 99.99\%
\][/tex]
- [tex]\[
\text{Percentage of students failing to satisfy the admission requirement} = 1 - 0.9999 = 0.0001 \text{ (or } 0.01\%)
\][/tex]
Therefore, approximately 0.0143% (or just a very small fraction) of students score below the admission requirement of 1148.
The question asks us to find the percentage of students from a high school who score below 1148 on their SATs, knowing that the scores are normally distributed with a mean of 96 and a standard deviation of 290.
To solve this, follow these steps:
1. Understand the Normal Distribution and Z-Score:
- The normal distribution is symmetrically centered around the mean.
- The Z-score tells us how many standard deviations an element is from the mean.
2. Calculate the Z-Score:
- The formula for the Z-score is:
[tex]\[
Z = \frac{X - \mu}{\sigma}
\][/tex]
where [tex]\(X\)[/tex] is the score we are interested in (1148), [tex]\(\mu\)[/tex] is the mean (96), and [tex]\(\sigma\)[/tex] is the standard deviation (290).
3. Substitute the values into the Z-score formula:
- [tex]\[
Z = \frac{1148 - 96}{290} \approx 3.63
\][/tex]
4. Use the Z-Score to Find the Cumulative Probability:
- The cumulative probability associated with a Z-score of 3.63 is approximately 0.9999 (looking this up in standard normal distribution tables or using a calculator).
5. Calculate the Remaining Percentage:
- Since 0.9999 (or 99.99%) of the students score below 1148, we need to subtract this from 1 to find the percentage that meet this requirement.
- [tex]\[
\text{Percentage of students scoring below 1148} = 99.99\%
\][/tex]
- [tex]\[
\text{Percentage of students failing to satisfy the admission requirement} = 1 - 0.9999 = 0.0001 \text{ (or } 0.01\%)
\][/tex]
Therefore, approximately 0.0143% (or just a very small fraction) of students score below the admission requirement of 1148.
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