Maria and Zoe are taking Biology 105 but are in different classes. Maria's class has an average of 78% with a standard deviation of 5% on the midterm, whereas Zoe's class has an average of 83% with a standard deviation of 12%. Assume that scores in both classes follow a normal distribution.
A. Convert Maria's midterm score of 84 to a standard z score.
which is correct?
a. 0.083
b. 0.5
c. 0.2
d. 1.2
e. 6
B Convert Zoe's midterm score of 89 to a standard z score.
a. 1.2
b. 0.5
c. 6
d. 0.917
e. 2.2
C. Who did better relative to her class?
a. Maria
b. Zoe
c. They performed the same.
d. Neither
e. Cannot determine
I will do the Maria part, then you do the rest
Mean = 78
score = 84
score above mean = 84 - 78 = 6
sigma = 5
6/5 = 1.2 standard deviations above mean
To calculate the z-score, we will use the formula:
z = (x - μ) / σ
where:
x = the value we want to convert to a z-score (midterm score)
μ = the mean of the class (average)
σ = the standard deviation of the class
A. Convert Maria's midterm score of 84 to a standard z score:
Given:
Maria's class: Average = 78%, Standard Deviation = 5%
Using the formula:
z = (x - μ) / σ
z = (84 - 78) / 5
z = 6 / 5
z = 1.2
So, the correct answer is d. 1.2
B. Convert Zoe's midterm score of 89 to a standard z score:
Given:
Zoe's class: Average = 83%, Standard Deviation = 12%
Using the formula:
z = (x - μ) / σ
z = (89 - 83) / 12
z = 6 / 12
z = 0.5
So, the correct answer is b. 0.5
C. To determine who did better relative to her class, we compare the z-scores. Since higher z-scores indicate a higher score relative to the class, we can conclude that the person with the higher z-score performed better relative to their class.
In this case, Zoe has a higher z-score (0.5) compared to Maria's z-score (1.2). Therefore, Zoe performed better relative to her class.
The correct answer is b. Zoe.