The distribution of scores on the SAT is approx. normal with mu= 500 and std dev=100.
a)what proportion of the population have SAT scores above 650?
b)what proportion of of the population have SAT scores below 540?
c)what is the minimum SAT score needed to be in the highest 20% of the population?
d)what SATE scores separates the top 60% from the rest of the distribution?
To solve these questions, we can use the standard normal distribution table or the Z-score formula. The Z-score formula allows us to convert any value from a normal distribution to a standard normal distribution.
a) To find the proportion of the population with SAT scores above 650, we need to calculate the Z-score and find the corresponding area under the standard normal curve.
Z = (X - μ) / σ
Z = (650 - 500) / 100
Z = 1.5
Using the standard normal distribution table or a Z-score calculator, we find that the area to the right of Z = 1.5 is approximately 0.0668. Therefore, approximately 0.0668 or 6.68% of the population have SAT scores above 650.
b) To find the proportion of the population with SAT scores below 540, we can use the Z-score formula.
Z = (X - μ) / σ
Z = (540 - 500) / 100
Z = 0.4
Using the standard normal distribution table or a Z-score calculator, we find that the area to the left of Z = 0.4 is approximately 0.6554. Therefore, approximately 0.6554 or 65.54% of the population have SAT scores below 540.
c) To find the minimum SAT score needed to be in the highest 20% of the population, we need to find the Z-score corresponding to the area of 0.20.
Using the standard normal distribution table or a Z-score calculator, we find that the Z-score for an area of 0.20 is approximately 0.84.
Z = (X - μ) / σ
0.84 = (X - 500) / 100
84 = X - 500
X = 584
Therefore, the minimum SAT score needed to be in the highest 20% of the population is 584.
d) To find the SAT scores that separate the top 60% from the rest of the distribution, we need to find the Z-score corresponding to the area of 0.60.
Using the standard normal distribution table or a Z-score calculator, we find that the Z-score for an area of 0.60 is approximately 0.25.
Z = (X - μ) / σ
0.25 = (X - 500) / 100
25 = X - 500
X = 525
Therefore, the SAT scores that separate the top 60% from the rest of the distribution are 525.
To answer these questions, we can use the concept of the standard normal distribution, also known as the Z-distribution. This distribution has a mean of 0 and a standard deviation of 1. By standardizing our values using the formula Z = (X - μ) / σ, we can calculate the proportion of scores above or below a certain value.
a) To find the proportion of the population with SAT scores above 650, we need to calculate the Z-score for 650. Using the formula, Z = (X - μ) / σ, we have Z = (650 - 500) / 100 = 1.5.
We can then consult the standard normal distribution table (often available online) or use statistical software to find the proportion associated with a Z-score of 1.5. In this case, the proportion is approximately 0.0668 or 6.68%.
Therefore, approximately 6.68% of the population will have SAT scores above 650.
b) Similarly, to find the proportion of the population with SAT scores below 540, we calculate the Z-score for 540: Z = (540 - 500) / 100 = 0.4.
Using the standard normal distribution table or software, we can find the proportion associated with a Z-score of 0.4. In this case, the proportion is approximately 0.6554 or 65.54%.
Therefore, approximately 65.54% of the population will have SAT scores below 540.
c) To find the minimum SAT score needed to be in the highest 20% of the population, we need to find the Z-score that corresponds to the 80th percentile. Since the standard normal distribution is symmetric, the 80th percentile is equivalent to 1 standard deviation below the mean.
Using the standard normal distribution table or software, we find that a Z-score of approximately 0.84 corresponds to the 80th percentile.
To convert this Z-score back to an SAT score, we use the formula X = Z * σ + μ:
X = 0.84 * 100 + 500 = 584.
Therefore, the minimum SAT score needed to be in the highest 20% of the population is 584.
d) To find the SAT score that separates the top 60% from the rest of the distribution, we need to find the Z-score that corresponds to the 40th percentile. Again, since the distribution is symmetric, the 40th percentile is equivalent to -0.25 standard deviations from the mean.
Using the standard normal distribution table or software, we find that a Z-score of approximately -0.25 corresponds to the 40th percentile.
Using the formula X = Z * σ + μ, we can find the SAT score:
X = -0.25 * 100 + 500 = 475.
Therefore, an SAT score of 475 separates the top 60% from the rest of the distribution.
Use the Z score formula:
Z = (X - ƒÊ)/SD
After finding the appropriate Z scores in a table in the back of your statistics text called something like "areas under the normal distribution," insert that Z value into the formula to find the value of your score.
Some problems need you to determine the Z score before consulting this table to get a proprotion.
I hope this helps. Thanks for asking.