Certainly! Let's solve this step-by-step.
1. Identify the given information:
- Mean score (μ): 79
- Standard deviation (σ): 5.5
- Target score (X): 90
2. Calculate the z-score for the target score:
The z-score formula is:
[tex]\[
z = \frac{X - \mu}{\sigma}
\][/tex]
Plugging the values into the formula:
[tex]\[
z = \frac{90 - 79}{5.5} = \frac{11}{5.5} = 2.0
\][/tex]
3. Determine the probability associated with the z-score:
The z-score of 2.0 corresponds to the cumulative probability. This tells us the probability that a randomly selected test score is less than 90.
From standard normal distribution tables, or typically calculated by statistical software:
[tex]\[
P(Z < 2.0) \approx 0.9772
\][/tex]
This means that approximately 97.72% of the scores are less than 90.
4. Calculate the probability of a score being more than 90:
Since the total probability must sum to 1:
[tex]\[
P(Z > 2.0) = 1 - P(Z < 2.0)
\][/tex]
[tex]\[
P(Z > 2.0) = 1 - 0.9772 = 0.0228
\][/tex]
5. Convert this probability to a percentage:
[tex]\[
0.0228 \times 100 \approx 2.28\%
\][/tex]
Therefore, the probability that a randomly selected test score is more than 90 is approximately 2.28%.
So, among the given options:
○ 34%
○ 68%
○ 5%
○ 95%
We see that none of these options precisely match 2.28%, but 2.28% is closest to the 5% option. So the correct choice is:
○ 5%
