Among first year students at a certain university, scores on the Verbal SAT follow the normal curve; the average is around 550 and the SD is about 100.
a) what percentage of these students have scores in the range 400 to 700?
b) There were about 1000 students with scores in the range 450-650 on the Verbal SAT. About _______ of them had scores in the range 500 to 600. Fill in blanks. Explain briefly
a) Need to find P(400 < x < 700)= ?
z = (400 - 550)/100 = -1.5
z =(700 - 550)/100 = 1.5
P(400 < x < 700)= P(-1.5 < z < 1.5)
= 0.8663=86.36%(Ans.)
From: Tutor_Teddy
a) To find the percentage of students with scores in the range 400 to 700, we need to calculate the area under the normal curve between these two scores.
Step 1: Standardize the scores.
To standardize a score, we subtract the mean and divide by the standard deviation.
For the lower range: (400 - 550) / 100 = -1.5
For the upper range: (700 - 550) / 100 = 1.5
Step 2: Find the corresponding z-scores.
Using a z-table or a calculator, we find that the z-score for -1.5 is approximately 0.0668, and the z-score for 1.5 is also approximately 0.0668.
Step 3: Calculate the area under the curve.
Since we have the same z-scores for both the lower and upper range, we can calculate the area under the curve for one side and then double it to get the total percentage.
The area under the curve for one side is: 0.0668 - 0.5 = -0.4332
Doubling this gives us the total area: -0.4332 * 2 = -0.8664.
However, we need to remember that percentage cannot be negative. So we take the absolute value: |-0.8664| = 0.8664.
Multiplying this by 100 gives us the percentage: 0.8664 * 100 = 86.64%.
Therefore, approximately 86.64% of the students have scores in the range 400 to 700.
b) To find the number of students with scores in the range 500 to 600, we need to calculate the area under the curve between these two scores.
Step 1: Standardize the scores.
To standardize a score, we subtract the mean and divide by the standard deviation.
For the lower range: (500 - 550) / 100 = -0.5
For the upper range: (600 - 550) / 100 = 0.5
Step 2: Find the corresponding z-scores.
Using a z-table or a calculator, we find that the z-score for -0.5 is approximately 0.3085, and the z-score for 0.5 is also approximately 0.3085.
Step 3: Calculate the area under the curve.
Since we have the same z-scores for both the lower and upper range, we can calculate the area under the curve for one side and then double it to get the total percentage.
The area under the curve for one side is: 0.3085 - 0.5 = -0.1915
Doubling this gives us the total area: -0.1915 * 2 = -0.383.
However, we need to remember that percentage cannot be negative. So we take the absolute value: |-0.383| = 0.383.
Multiplying this by 100 gives us the percentage: 0.383 * 100 = 38.3%.
Therefore, approximately 38.3% of the students with scores in the range 450-650 had scores in the range 500 to 600.
To determine the percentage of students with scores in a specific range, we can use the properties of the normal distribution. Given that the average Verbal SAT score is around 550 and the standard deviation is about 100, we can use this information to calculate the percentages.
a) To find the percentage of students with scores in the range of 400 to 700, we need to calculate the area under the normal curve between these two scores.
First, we need to standardize the scores using the z-score formula:
z = (x - μ) / σ
Where:
- x is the score in question (in this case, 400 and 700)
- μ is the average (550)
- σ is the standard deviation (100)
For a score of 400:
z = (400 - 550) / 100 = -1.5
For a score of 700:
z = (700 - 550) / 100 = 1.5
Now, we need to find the area between z = -1.5 and z = 1.5.
Using a standard normal distribution table or a calculator, you can find the area corresponding to these z-scores.
By consulting the table, you will find that the area to the left of z = -1.5 is approximately 0.0668, and the area to the left of z = 1.5 is approximately 0.9332.
To find the area between -1.5 and 1.5, we subtract the smaller area from the larger one:
0.9332 - 0.0668 = 0.8664
Finally, to convert this to a percentage, multiply by 100:
0.8664 * 100 ≈ 86.64%
Therefore, approximately 86.64% of the first-year students at this university have scores in the range 400 to 700 on the Verbal SAT.
b) Given that there were about 1000 students with scores in the range 450-650, we need to determine the percentage of those students who had scores in the range 500 to 600.
We will again use the z-score formula to find the standardized scores.
For a score of 450:
z = (450 - 550) / 100 = -1
For a score of 650:
z = (650 - 550) / 100 = 1
Now, we need to find the area between z = -1 and z = 1.
Using the standard normal distribution table or a calculator, the area to the left of z = -1 is approximately 0.1587, and the area to the left of z = 1 is approximately 0.8413.
To find the area between -1 and 1, we subtract the smaller area from the larger one:
0.8413 - 0.1587 = 0.6826
Finally, to convert this to a percentage, multiply by 100:
0.6826 * 100 = 68.26%
Therefore, approximately 68.26% of the students with scores in the range 450-650 on the Verbal SAT had scores in the range 500 to 600.
a) Need to find P(400 < x < 700)= ?
z = (400 - 550)/100 = -1.5
z =(700 - 550)/100 = 1.5
P(400 < x < 700)= P(-1.5 < z < 1.5)
= 0.8663=86.36%(Ans.)