Scores on the SAT verbal test in recent years follow approximately N(504,111) distribution. What is the proportion of students scoring between 450 and 550? How high must a student score to place in the top 10% of all students taking the SAT? Please show the steps to obtaining this answer.

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http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html
(a) Proportion between 450 and 550: 34.7%
(b) Top 10% cutoff score: 646

scores on the sat verbal test in recent years follow approximately the n(504,111) distribution. the proportion of students between 450 and 550 is

To find the proportion of students scoring between 450 and 550, we need to calculate the area under the normal curve between these two scores.

Step 1: Convert the scores to standard units
We first need to convert the scores 450 and 550 into standard units using the formula:
Standard Unit = (Observed Value - Mean) / Standard Deviation

For the score of 450:
Standard unit = (450 - 504) / 111 ≈ -0.486

For the score of 550:
Standard unit = (550 - 504) / 111 ≈ 0.414

Step 2: Look up the z-scores
Next, we will look up the corresponding z-scores for these standard units in the standard normal distribution table. The table gives the proportion of values below a given z-score.

Looking up the z-score of -0.486, we find that the proportion of values below this z-score is approximately 0.3149.

Looking up the z-score of 0.414, we find that the proportion of values below this z-score is approximately 0.6591.

Step 3: Calculate the proportion between the scores
To find the proportion of students scoring between 450 and 550, we need to find the difference between the proportion below the upper score and the proportion below the lower score:

Proportion = Proportion (upper score) - Proportion (lower score)
= 0.6591 - 0.3149
≈ 0.3442

Therefore, approximately 34.42% of students score between 450 and 550 on the SAT verbal test.

To determine the score needed to place in the top 10% of all students, we need to find the z-score that corresponds to the 90th percentile.

Step 4: Find the z-score for the 90th percentile
The z-score that corresponds to the 90th percentile can be found by looking up the value in the standard normal distribution table. The z-score for the 90th percentile is approximately 1.282.

Step 5: Convert the z-score to an actual score
Using the z-score formula, we can solve for the actual score:

Standard Unit = (Observed Value - Mean) / Standard Deviation
1.282 = (Observed Value - 504) / 111

Solving for the observed value:
111 * 1.282 = Observed Value - 504
Observed Value = (111 * 1.282) + 504

Observed Value ≈ 657.51

Therefore, a student must score approximately 657.51 or higher to place in the top 10% of all students taking the SAT.

To find the proportion of students scoring between 450 and 550 on the SAT verbal test, we need to calculate the area under the normal distribution curve between these two scores.

Step 1: Standardize the scores
We need to convert the scores 450 and 550 into standardized z-scores using the formula:

z = (x - μ) / σ

Where:
x = individual score
μ = mean of the distribution
σ = standard deviation of the distribution

Given that the mean (μ) is 504 and the standard deviation (σ) is 111, we can calculate the z-scores for 450 and 550:

z1 = (450 - 504) / 111
z2 = (550 - 504) / 111

Step 2: Look up the area in the standard normal distribution table
Once we have the z-scores, we can look up the corresponding areas in the standard normal distribution table. The table gives us the proportion of scores falling below a specific z-score.

For z-score 450:
We need to look up the area to the left of z1 in the standard normal distribution table. Let's say this value is A1.

For z-score 550:
We need to look up the area to the left of z2 in the standard normal distribution table. Let's say this value is A2.

Step 3: Calculate the proportion between 450 and 550
To find the proportion of students scoring between 450 and 550, we subtract A1 from A2:

Proportion = A2 - A1

Now let's move on to the second part of the question, determining the score required to place in the top 10% of all students.

To find this score, we need to determine the z-score that corresponds to the top 10% of the distribution.

Step 4: Find the z-score for the top 10% area
We want to find the z-score that leaves 10% of the area to the right under the normal distribution curve. In other words, we want to find the z-score that has an area of 0.10 to its right.

From the standard normal distribution table, find the z-score that corresponds to an area of 0.10. Let's say this z-score is Z.

Step 5: Convert the z-score to the actual score
Using the formula for z-scores, we can find the actual score corresponding to the z-score Z:

score = μ + Z * σ

Substitute the values of μ and σ from the given information to calculate the answer.

The proportion of students scoring between 450 and 550 can be found by subtracting the area to the left of z1 from the area to the left of z2. To find the score required to place in the top 10% of all students, we need to determine the z-score corresponding to an area of 0.10 to the right under the normal distribution curve, and then convert this z-score to the actual score.