In a certain school district in a large metropolitan area, the SAT scores over that past five years are normally distributed with a mean of 1476. Furthermore, P20 is 1209.
For a normal distribution, what is the P99 score for this population?
To identify the P99 score, which is the value below which 99% of the scores fall, we need to know the standard deviation of the scores in this particular distribution. Since we know that the distribution is normal and we have been given the P20 (the 20th percentile), we can use this information to estimate the standard deviation.
The 20th percentile means that 20% of the scores fall below this number. In a normal distribution, the percentile rank corresponds to a specific z-score. The z-score for P20 can be looked up in a standard normal distribution table or found using a statistical software package or calculator that outputs z-scores based upon percentiles.
The z-score corresponding to the 20th percentile (P20) is typically around -0.8416. The equation to convert a z-score to an actual score in the distribution is:
X = μ + Z(σ)
where X is the actual score (in this case, 1209 for P20), μ is the mean (1476), Z is the z-score (-0.8416 for P20), and σ is the standard deviation.
We can rearrange this formula to solve for the standard deviation (σ):
σ = (X - μ) / Z
Insert the values for P20:
σ = (1209 - 1476) / -0.8416
σ ≈ (267) / 0.8416
σ ≈ 317.36598
Now let's find the z-score corresponding to P99 (the 99th percentile). This z-score is typically around 2.3263.
Then we calculate the score corresponding to this z-score using the original formula:
X = μ + Z(σ)
X ≈ 1476 + (2.3263)(317.36598)
X ≈ 1476 + 738.383
X ≈ 2214.383
Therefore, the P99 score for this population is approximately 2214, rounding to the nearest whole number if not dealing with decimals. Keep in mind that these values are estimates as they rely on the 20th percentile to estimate the standard deviation and the precise z-scores are used from statistical tables or computational tools.