Find the score that separates the top 75% from the bottom 25% if the Mean is 100 and the Standard deviation is 15
See later post.
To find the score that separates the top 75% from the bottom 25%, we need to calculate the z-score corresponding to the desired percentile and then convert it back to the original score.
1. Identify the z-score corresponding to the desired percentile:
- The top 75% corresponds to the percentile rank of 100 - 75 = 25.
- Using a standard normal distribution table or a calculator, we find that the z-score corresponding to a percentile rank of 25 is approximately -0.674.
2. Convert the z-score back to the original score:
- The formula to convert a z-score to an original score is: original score = (z-score * standard deviation) + mean.
- In this case, substituting -0.674 for the z-score, 15 for the standard deviation, and 100 for the mean:
original score = (-0.674 * 15) + 100
= -10.11 + 100
= 89.89
Therefore, the score that separates the top 75% from the bottom 25% is approximately 89.89.
To find the score that separates the top 75% from the bottom 25%, we can make use of the concept of z-scores. A z-score measures the standard deviations away from the mean a particular value is.
First, we need to find the z-score corresponding to the top 75% (or the 75th percentile). We can use a standard normal distribution table or a calculator to find this value. The z-score that corresponds to the top 75% is approximately 0.674.
Next, we use the formula for z-score, which is:
z-score = (x - mean) / standard deviation
Rearranging the formula, we can solve for x:
x = (z-score * standard deviation) + mean
Plugging in the values:
x = (0.674 * 15) + 100
Calculating this equation, we find:
x ≈ 110.11
Therefore, the score that separates the top 75% from the bottom 25% is approximately 110.11.