On a test whose distribution is approximately normal with a mean of 50 and a standard deviation of 10, the results for three students were reported as follows:
Student Opie has a T-score of 60.
Student Paul has a z-score of -1.00.
Student Quincy has a z-score of +2.00.
Obtain the z-score and T-score for EACH student. Show your calculations. Who did better on the test? How many standard deviation units is each score from the mean? Compare the results of the three students.
T = 50 + 10Z
Z = your score in terms of standard deviations.
With this information, you should be able to answer the questions.
To obtain the z-score for each student, we can use the formula:
z = (x - μ) / σ
Where
x = student's score
μ = mean
σ = standard deviation
For Student Opie:
z = (x - μ) / σ
= (60 - 50) / 10
= 10 / 10
= 1.00
To convert the z-score to a T-score, we use the following formula:
T = (10 * z) + 50
T = (10 * 1.00) + 50
= 10 + 50
= 60
For Student Paul:
Given that Paul's score has already been provided as a z-score of -1.00
To find the T-score, we can use the formula:
T = (10 * z) + 50
= (10 * -1) + 50
= -10 + 50
= 40
For Student Quincy:
Given that Quincy's score has already been provided as a z-score of +2.00
To find the T-score, we can use the formula:
T = (10 * z) + 50
= (10 * 2) + 50
= 20 + 50
= 70
Now let's compare the results of the three students:
Opie's T-score: 60
Paul's T-score: 40
Quincy's T-score: 70
The student with the highest T-score, Quincy, did better on the test.
To determine how many standard deviation units each score is from the mean:
Opie's z-score: 1.00
Paul's z-score: -1.00
Quincy's z-score: 2.00
Opie's score is 1 standard deviation unit above the mean.
Paul's score is 1 standard deviation unit below the mean.
Quincy's score is 2 standard deviation units above the mean.
To find the z-score and T-score for each student, we can use the following formulas:
Z-score = (x - μ) / σ
T-score = (x - μ) / (σ * (1 / n))
Where:
x = student's score
μ = mean of the distribution
σ = standard deviation of the distribution
n = sample size (typically the population size)
Now, let's calculate the z-score and T-score for each student and compare the results:
1. Student Opie:
Z-score = (60 - 50) / 10 = 1
T-score = (60 - 50) / (10 * (1 / 1)) = 1
Opie's z-score and T-score are both 1.
2. Student Paul:
Z-score = (-1.00) --> The given z-score is already provided for Paul.
T-score = (-1.00 - 0) / (10 * (1 / 1)) = -1.00
Paul's z-score is -1.00 and T-score is also -1.
3. Student Quincy:
Z-score = (+2.00) --> The given z-score is already provided for Quincy.
T-score = (+2.00 - 0) / (10 * (1 / 1)) = 2.00
Quincy's z-score is +2.00 and T-score is also 2.
Now let's compare the results:
Opie has a z-score of 1 and T-score of 1.
Paul has a z-score of -1 and T-score of -1.
Quincy has a z-score of +2 and T-score of 2.
To determine who did better on the test, we need to consider that T-scores are more appropriate when comparing different sample sizes. Since all three students have the same sample size of 1, we can directly compare their z-scores.
Quincy (z-score of +2) performed better on the test compared to all, as a higher z-score indicates a higher relative position in the distribution. Opie (z-score of 1) performed better than Paul (z-score of -1) but not as well as Quincy.
To determine how many standard deviation units each score is from the mean, we can look at the absolute value of the z-scores:
Opie: 1 standard deviation above the mean.
Paul: 1 standard deviation below the mean.
Quincy: 2 standard deviations above the mean.
Thus, Quincy performed the best with a z-score of +2, followed by Opie with a z-score of +1, and Paul performed the worst with a z-score of -1.