Sure, let's break down the problem step-by-step:
1. Understanding the Problem:
We have a set of exam scores that follow a normal distribution with:
- Mean ([tex]\(\mu\)[/tex]) = 60 points
- Standard Deviation ([tex]\(\sigma\)[/tex]) = 6 points
Anita's exam score is 51 points. We need to find the proportion of exam scores that are higher than 51 points.
2. Standardize Anita’s Score:
To understand where Anita's score falls within the distribution, we convert her raw score into a z-score. The z-score measures how many standard deviations a data point is from the mean.
[tex]\[
z = \frac{{X - \mu}}{{\sigma}}
\][/tex]
Plugging in the values:
[tex]\[
z = \frac{{51 - 60}}{{6}} = \frac{{-9}}{{6}} = -1.5
\][/tex]
3. Determine Cumulative Proportion:
The z-score of -1.5 tells us how Anita's score compares to the rest of the distribution. We will use the cumulative distribution function (CDF) of the standard normal distribution to find the proportion of scores less than Anita’s score.
Looking up the CDF value for a z-score of -1.5, we find:
[tex]\[
\Phi(-1.5) = 0.0668
\][/tex]
This means that approximately 6.68% of the scores are less than Anita’s score.
4. Find the Proportion Higher than Anita’s Score:
Since the total area under the normal distribution curve represents 100% of the scores, the proportion of scores that are higher than Anita’s score is:
[tex]\[
1 - \Phi(-1.5) = 1 - 0.0668 = 0.9332
\][/tex]
5. Conclusion:
Thus, the proportion of exam scores that are higher than Anita's score is:
[tex]\[
0.9332
\][/tex]
So, the proportion of exam scores higher than Anita's score is 0.9332 (rounded to four decimal places).
