Answer:
a)
Kyle's z-score was 1.9 to the nearest tenth
Ethan's z-score was -0.8 to the nearest tenth
b)
The percent of the students had a final exam score lower than Ethan's score was 21.19%
Explanation:
a) Lets revise how to find the z-score
- The rule the z-score is z = (x - μ)/σ , where
# x is the score
# μ is the mean
# σ is the standard deviation
* Lets solve the problem
- The exam scores are normally distributed with a mean of 105 and a
standard deviation of 16
∴ μ = 105 and σ = 16
- Kyle and Ethan are took the final exam
- Kyle's score was 135
- Ethan's score was 93
- Lets find the z-score for each one
∵ Kyle's score was 135
∴ x = 135
∵ μ = 105 and σ = 16
∵ z-score = (x - μ)/σ
∴ z-score for Kyle = (135 - 105)/16 = 30/16 = 15/8 = 1.875
* Kyle's z-score is 1.9 to the nearest tenth
∵ Ethan's score was 93
∴ x = 93
∵ μ = 105 and σ = 16
∵ z-score = (x - μ)/σ
∴ z-score for Ethan = (93 - 105)/16 = -12/16 = -3/4 = -0.75
* Ethan's z-score is -0.8 to the nearest tenth
b) To find the percent of students with a lower exam score than Ethan
you will asking to find the proportion of area under the standard
normal distribution curve for all z-scores < -0.8
- It can be read from a z-score table by referencing a z-score of -0.8
- Look to the attached file
∴ The value from the table is 0.2119
- To change it to percent multiply it by 100%
∴ 0.2119 × 100% = 21.19%
* The percent of the students had a final exam score lower than
Ethan's score was 21.19%