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 z-score for the 99-th percentile?
To find the z-score for the 99th percentile of a normally distributed SAT score, we need to calculate the z-score from the standard normal distribution that corresponds to the 99th percentile.
The z-score formula for any value X in a normal distribution is:
\[ z = \frac{X - \mu}{\sigma} \]
where:
- \( z \) is the z-score,
- \( X \) is the value from the normal distribution,
- \( \mu \) is the mean of the distribution, and
- \( \sigma \) is the standard deviation of the distribution.
Since P20 is the 20th percentile (the value below which 20% of the data fall), and you've mentioned that P20 is 1209, this corresponds to a z-score of -0.8416 (which is the z-score for the 20th percentile as per the standard normal distribution table).
Now we can use this information to calculate the standard deviation of the SAT scores:
\[ Z_{P20} = \frac{X_{P20} - \mu}{\sigma} \]
\[ -0.8416 = \frac{1209 - 1476}{\sigma} \]
\[ \sigma = \frac{1209 - 1476}{-0.8416} \]
\[ \sigma = \frac{-267}{-0.8416} \]
\[ \sigma \approx 317.2 \]
Now that we have \( \sigma \), we can find the value that corresponds to the 99th percentile (which we will label as \( X_{99} \)). You can look up the z-score for the 99th percentile in standard normal distribution tables or use a statistical software or calculator. The z-score for the 99th percentile is approximately 2.33.
We can then use this z-score to find \( X_{99} \) as follows:
\[ Z_{99} = \frac{X_{99} - \mu}{\sigma} \]
\[ 2.33 = \frac{X_{99} - 1476}{317.2} \]
\[ X_{99} - 1476 = 2.33 \cdot 317.2 \]
\[ X_{99} - 1476 = 739.196 \]
\[ X_{99} = 739.196 + 1476 \]
\[ X_{99} \approx 2215.2 \]
Since your question didn't ask for the actual SAT score but the z-score corresponding to the 99th percentile, the z-score is approximately 2.33.