A normal distribution has a mean of 120 and a standard deviation of 20. What score separates the top 40% of the scores from the rest?

Z = (score-mean)/SD

Look in the back of your statistics textbook for a table called something like “area under normal distribution” to find the proportion/probability (.40) and its Z score. Insert data into above equation and solve for score.

Well, if the top 40% of the scores are separating from the rest, I guess you could say it's like a high school popularity contest. So, in this case, the score that separates the cool kids from the not-so-cool kids would be ... drum roll, please ... 130! It's like the class clown getting the last laugh.

To find the score that separates the top 40% of the scores from the rest in a normal distribution with a mean of 120 and a standard deviation of 20, you can use the Z-score formula.

1. Start by finding the Z-score associated with the top 40% of the distribution. To do this, subtract the percentage (40%) from 100% to find the percentage in the lower tail of the distribution. The lower tail percentage is 100% - 40% = 60%.

2. Convert the lower tail percentage to a Z-score using a Z-table or a calculator. The Z-score associated with a lower tail percentage of 60% is approximately 0.2533.

3. Since we want to find the score that separates the top 40% of the distribution, we need to find the raw score corresponding to the Z-score calculated in step 2. To find the raw score, use the Z-score formula: X = Z * σ + μ, where X is the raw score, Z is the Z-score, σ is the standard deviation, and μ is the mean.

Plugging in the values, X = 0.2533 * 20 + 120 = 125.07.

Therefore, the score that separates the top 40% of the scores from the rest is approximately 125.07.

To find the score that separates the top 40% from the rest in a normal distribution with a known mean and standard deviation, you can use the z-score formula and the z-table.

1. Calculate the z-score corresponding to the desired percentile (in this case, the top 40% or 0.40). The z-score represents the number of standard deviations away from the mean, and we can find it using the standard normal distribution table or calculator.

2. Locate the z-score in the z-table. The z-table provides the area (or percentile) to the left of a given z-score. Find the closest z-score that matches or is slightly greater than the calculated z-score. Note down the corresponding area.

3. Subtract the area obtained from 0.50 (which represents 50%) to get the area to the right of the z-score.

4. Use the z-score formula to find the actual score on the distribution, given the calculated z-score, mean, and standard deviation. The formula is: score = (z-score * standard deviation) + mean.

Let's go through the calculation steps:

1. The z-score corresponding to the top 40% is obtained by subtracting 0.40 from 1.00 (since we need the area to the right). So, 1.00 - 0.40 = 0.60.

2. Looking up a z-table or using a calculator, you will find that a z-score of approximately 0.25 corresponds to an area of 0.5987, which is slightly greater than 0.60.

3. Subtracting 0.5987 from 0.50, we get 0.50 - 0.5987 = 0.10. This represents the area to the right of the desired z-score.

4. Now, using the z-score formula, we can find the actual score. Plugging in the known values: z-score = 0.25, mean = 120, and standard deviation = 20.

score = (0.25 * 20) + 120
= 5 + 120
= 125

Therefore, the score that separates the top 40% from the rest is 125.

125