To determine how many standard deviations each score is from the mean, we will use the concept of the z-score. A z-score measures how many standard deviations an element is from the mean. The formula for the z-score is:
[tex]\[ z = \frac{X - \mu}{\sigma} \][/tex]
where:
- [tex]\(X\)[/tex] is the individual score.
- [tex]\(\mu\)[/tex] is the mean of the scores.
- [tex]\(\sigma\)[/tex] is the standard deviation.
Given:
- [tex]\(\mu = 216\)[/tex]
- [tex]\(\sigma = 52\)[/tex]
Let's calculate the z-score for each of the given scores.
### For the score 253:
[tex]\[ z = \frac{253 - 216}{52} \][/tex]
[tex]\[ z = \frac{37}{52} \][/tex]
[tex]\[ z \approx 0.71 \][/tex]
### For the score 286:
[tex]\[ z = \frac{286 - 216}{52} \][/tex]
[tex]\[ z = \frac{70}{52} \][/tex]
[tex]\[ z \approx 1.35 \][/tex]
### For the score 300:
[tex]\[ z = \frac{300 - 216}{52} \][/tex]
[tex]\[ z = \frac{84}{52} \][/tex]
[tex]\[ z \approx 1.62 \][/tex]
### For the score 86:
[tex]\[ z = \frac{86 - 216}{52} \][/tex]
[tex]\[ z = \frac{-130}{52} \][/tex]
[tex]\[ z \approx -2.50 \][/tex]
### Summary
Each score is within the following number of standard deviations from the mean:
- The score 253 is approximately [tex]\(0.71\)[/tex] standard deviations above the mean.
- The score 286 is approximately [tex]\(1.35\)[/tex] standard deviations above the mean.
- The score 300 is approximately [tex]\(1.62\)[/tex] standard deviations above the mean.
- The score 86 is approximately [tex]\(-2.50\)[/tex] standard deviations below the mean.
