Question What percentage of students from a high school with normally distributed SAT scores (mean $1497$ , std. dev. $310$ ) meet the college's minimum score requirement of $536$ ?

Studdy Solution

STEP 1

Assumptions
1. The SAT scores are normally distributed.
2. The mean (average) SAT score is 1497.
3. The standard deviation of the SAT scores is 310.
4. The minimum SAT score required for college admission is 536.
5. We are looking for the percentage of students scoring at or above 536.

STEP 2

To find the percentage of students who score at or above the minimum requirement, we need to calculate the z-score for the score of 536. The z-score is a measure of how many standard deviations an element is from the mean.
$z = \frac{{X - \mu}}{{\sigma}}$
Where: - $X$ is the score of interest (536 in this case), - $\mu$ is the mean of the distribution (1497), - $\sigma$ is the standard deviation of the distribution (310).

STEP 3

Now, plug in the values for $X$ , $\mu$ , and $\sigma$ to calculate the z-score for the score of 536.
$z = \frac{{536 - 1497}}{{310}}$

STEP 4

Calculate the z-score.
$z = \frac{{536 - 1497}}{{310}} = \frac{{-961}}{{310}} \approx -3.10$

STEP 5

The z-score tells us how many standard deviations the score of 536 is below the mean. Now we need to find the percentage of students that score above this z-score.
We can use the standard normal distribution table or a calculator with normal distribution functions to find the probability that a student scores higher than a z-score of -3.10.

STEP 6

Using the standard normal distribution table or a calculator, we find the cumulative probability for a z-score of -3.10.
The cumulative probability represents the percentage of students scoring below 536. To find the percentage of students scoring at or above 536, we subtract this value from 1.

STEP 7

If the cumulative probability for a z-score of -3.10 is found to be $p$ , then the percentage of students scoring at or above 536 is:
$Percentage = (1 - p) \times 100\%$

STEP 8

Assuming the cumulative probability for a z-score of -3.10 is approximately 0.0010 (this value may vary slightly depending on the source of the normal distribution table or the precision of the calculator), we calculate the percentage of students meeting the admission requirement.
$Percentage = (1 - 0.0010) \times 100\%$

STEP 9

Calculate the percentage.
$Percentage = (0.9990) \times 100\% = 99.90\%$
Approximately 99.90% of students from this school earn scores that satisfy the admission requirement.

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Question What percentage of students from a high school with normally distributed SAT scores (mean $1497$ , std. dev. $310$ ) meet the college's minimum score requirement of $536$ ?

Studdy Solution

STEP 1

Assumptions
1. The SAT scores are normally distributed.
2. The mean (average) SAT score is 1497.
3. The standard deviation of the SAT scores is 310.
4. The minimum SAT score required for college admission is 536.
5. We are looking for the percentage of students scoring at or above 536.

STEP 2

STEP 3

Now, plug in the values for $X$ , $\mu$ , and $\sigma$ to calculate the z-score for the score of 536.
$z = \frac{{536 - 1497}}{{310}}$

STEP 4

Calculate the z-score.
$z = \frac{{536 - 1497}}{{310}} = \frac{{-961}}{{310}} \approx -3.10$

STEP 5

STEP 6

Using the standard normal distribution table or a calculator, we find the cumulative probability for a z-score of -3.10.
The cumulative probability represents the percentage of students scoring below 536. To find the percentage of students scoring at or above 536, we subtract this value from 1.

STEP 7

If the cumulative probability for a z-score of -3.10 is found to be $p$ , then the percentage of students scoring at or above 536 is:
$Percentage = (1 - p) \times 100\%$

STEP 8

STEP 9

Calculate the percentage.
$Percentage = (0.9990) \times 100\% = 99.90\%$
Approximately 99.90% of students from this school earn scores that satisfy the admission requirement.

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Question What percentage of students from a high school with normally distributed SAT scores (mean 149714971497, std. dev. 310310310) meet the college's minimum score requirement of 536536536?

Studdy Solution

STEP 1

Assumptions1. The SAT scores are normally distributed.2. The mean (average) SAT score is 1497.3. The standard deviation of the SAT scores is 310.4. The minimum SAT score required for college admission is 536.5. We are looking for the percentage of students scoring at or above 536.

STEP 2

STEP 3

Now, plug in the values for XXX, μ\muμ, and σ\sigmaσ to calculate the z-score for the score of 536.z=536−1497310z = \frac{{536 - 1497}}{{310}}z=310536−1497​

STEP 4

Calculate the z-score.z=536−1497310=−961310≈−3.10z = \frac{{536 - 1497}}{{310}} = \frac{{-961}}{{310}} \approx -3.10z=310536−1497​=310−961​≈−3.10

STEP 5

STEP 6

Using the standard normal distribution table or a calculator, we find the cumulative probability for a z-score of -3.10.The cumulative probability represents the percentage of students scoring below 536. To find the percentage of students scoring at or above 536, we subtract this value from 1.

STEP 7

If the cumulative probability for a z-score of -3.10 is found to be ppp, then the percentage of students scoring at or above 536 is:Percentage=(1−p)×100%Percentage = (1 - p) \times 100\%Percentage=(1−p)×100%

STEP 8

STEP 9

Calculate the percentage.Percentage=(0.9990)×100%=99.90%Percentage = (0.9990) \times 100\% = 99.90\%Percentage=(0.9990)×100%=99.90%Approximately 99.90% of students from this school earn scores that satisfy the admission requirement.

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Question What percentage of students from a high school with normally distributed SAT scores (mean $1497$ , std. dev. $310$ ) meet the college's minimum score requirement of $536$ ?

Assumptions
1. The SAT scores are normally distributed.
2. The mean (average) SAT score is 1497.
3. The standard deviation of the SAT scores is 310.
4. The minimum SAT score required for college admission is 536.
5. We are looking for the percentage of students scoring at or above 536.

Now, plug in the values for $X$ , $\mu$ , and $\sigma$ to calculate the z-score for the score of 536.
$z = \frac{{536 - 1497}}{{310}}$

Calculate the z-score.
$z = \frac{{536 - 1497}}{{310}} = \frac{{-961}}{{310}} \approx -3.10$

Using the standard normal distribution table or a calculator, we find the cumulative probability for a z-score of -3.10.
The cumulative probability represents the percentage of students scoring below 536. To find the percentage of students scoring at or above 536, we subtract this value from 1.

If the cumulative probability for a z-score of -3.10 is found to be $p$ , then the percentage of students scoring at or above 536 is:
$Percentage = (1 - p) \times 100\%$

Calculate the percentage.
$Percentage = (0.9990) \times 100\% = 99.90\%$
Approximately 99.90% of students from this school earn scores that satisfy the admission requirement.