every year thousands of students take the SAT'S. Recent trends show that the scores on the verbal section follow an approx. normal distribution with mean of 350 and stand. deviation of 110
A)a student is chosen at random. it is noted that only 8% of students have scored higher. what score did the student receive.
b)68.26% of students who take the SAT SCORE BETWEEN _____ AND______
95.44% OF STUDENTS WHO TAKE THE SAT SCORE BETWEEN _____AND_____
99.74% OS STUDENTS WHO TAKE THE SAT SCORE BETWEEN _____AND______
Idk how to do a but for part b just use the emperical formula i think? except the numbers r kinda funky....so for 68.26 it wud be between 240 and 460, for 95.44 it wud be 130 and 570, and for 99.74 it wud be 20 and 680!
To answer these questions, we need to use the concept of z-scores. A z-score allows us to standardize a data value with respect to the mean and standard deviation. It tells us how many standard deviations a particular data point is away from the mean.
Formally, the z-score is given by the formula: z = (x - μ) / σ, where x is the data value, μ is the mean, and σ is the standard deviation.
a) To find the score of a student who scored higher than 92% of students, we need to find the z-score that corresponds to that percentile.
Using the standard normal distribution table or a calculator, we can find the z-score that corresponds to 92%, which is approximately 1.405.
Substituting the values into the formula, we can solve for x:
1.405 = (x - 350) / 110
Now, we can solve for x:
1.405 * 110 = x - 350
154.55 + 350 = x
504.55 = x
So the student received a score of approximately 505.
b) To find the range of scores for the given percentile ranges, we can use the properties of the normal distribution.
For the first question, 68.26% of students will have scores within one standard deviation of the mean. So we need to find the z-scores that correspond to 34.13% of students above and below the mean.
Using the standard normal distribution table or a calculator, we find that a z-score of approximately -0.994 corresponds to 34.13%. We can use this value to find the score below the mean:
-0.994 = (x - 350) / 110
Solving for x:
-0.994 * 110 = x - 350
-109.34 + 350 = x
240.66 = x
So, the range of scores for 68.26% of students is approximately 240.66 to 459.34.
For the second question, 95.44% of students will have scores within two standard deviations of the mean. This means we need to find the z-scores that correspond to 47.72% of students above and below the mean.
Using the standard normal distribution table or a calculator, we find that a z-score of approximately -1.96 corresponds to 47.72%. We can use this value to find the score below the mean:
-1.96 = (x - 350) / 110
Solving for x:
-1.96 * 110 = x - 350
-215.6 + 350 = x
134.4 = x
Similarly, two standard deviations above the mean will be:
1.96 = (x - 350) / 110
Solving for x:
1.96 * 110 = x - 350
215.6 + 350 = x
565.6 = x
So, the range of scores for 95.44% of students is approximately 134.4 to 565.6.
For the last question, 99.74% of students will have scores within three standard deviations of the mean. This means we need to find the z-scores that correspond to 49.87% of students above and below the mean.
Using the standard normal distribution table or a calculator, we find that a z-score of approximately -2.97 corresponds to 49.87%. We can use this value to find the score below the mean:
-2.97 = (x - 350) / 110
Solving for x:
-2.97 * 110 = x - 350
-326.7 + 350 = x
23.3 = x
Similarly, three standard deviations above the mean will be:
2.97 = (x - 350) / 110
Solving for x:
2.97 * 110 = x - 350
326.7 + 350 = x
676.7 = x
So, the range of scores for 99.74% of students is approximately 23.3 to 676.7.