Maria and Juana recieved scores of 82 and 65, respectively. If these score are equivalent to z-scores of 1.6 and -0.2, respectively,what are the mean and the standard deviation of the scores of all the students who took the examination?
z = (x - m)/s, where s is the standard deviation and m is the mean.
so you end up with 2 equations in 2 unknowns
1.6 = (82-m)/s
1.6s = 82 - m -----> 10m + 16s = 820
-.2 = (65-m)/s
-.2s = 65 - m ---> 10m - 2s = 650
easy to solve for m and s
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To find the mean and standard deviation of the scores of all students who took the examination, we can use the z-score formula.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
z = z-score
x = raw score
μ = mean
σ = standard deviation
We are given the z-scores for Maria and Juana, which are 1.6 and -0.2 respectively. Let's denote μ as the mean and σ as the standard deviation.
For Maria:
1.6 = (82 - μ) / σ
For Juana:
-0.2 = (65 - μ) / σ
We have two equations with two unknowns. We can solve this system of equations to find the mean (μ) and standard deviation (σ).
Rearranging the first equation:
1.6σ = 82 - μ
Rearranging the second equation:
-0.2σ = 65 - μ
We can solve these equations simultaneously by using substitution or elimination method.
Let's use the elimination method:
Multiply the first equation by 0.2:
0.32σ = 16.4 - 0.2μ
Since we have -0.2σ in the second equation, we can add this equation to the previous equation:
0.32σ - 0.2σ = 16.4 - 0.2μ - 65 + μ
0.12σ = -48.6 + 0.8μ
Divide the equation by 0.12:
σ = (-48.6 + 0.8μ) / 0.12
Substitute this value back into the second equation:
-0.2((-48.6 + 0.8μ) / 0.12) = 65 - μ
Multiply both sides by 0.12 to eliminate the fraction:
-0.024(-48.6 + 0.8μ) = 7.8 - 0.12μ
1.1664 - 0.0192μ = 7.8 - 0.12μ
Rearrange the equation:
0.1008μ - 6.6336 = 0
0.1008μ = 6.6336
μ = 6.6336 / 0.1008
μ ≈ 65.83
Now we can substitute the value of μ into one of the original equations to find σ:
Using the first equation:
1.6 = (82 - 65.83) / σ
Rearrange the equation:
1.6σ = 16.17
σ = 16.17 / 1.6
σ ≈ 10.11
Therefore, the mean (μ) of all the student's scores is approximately 65.83, and the standard deviation (σ) is approximately 10.11.
To find the mean and standard deviation of the scores of all the students who took the examination, we can use the z-score formula.
The formula for z-score is:
z = (X - μ) / σ
Where:
z is the z-score,
X is the value,
μ is the population mean, and
σ is the population standard deviation.
Given that Maria's score has a z-score of 1.6 and Juana's score has a z-score of -0.2, we can use these values to find the mean and standard deviation.
For Maria:
1.6 = (82 - μ) / σ ----(1)
For Juana:
-0.2 = (65 - μ) / σ ----(2)
We have two equations with two unknowns (μ and σ). We can solve this system of equations to find the values of μ and σ.
Let's first find the population mean (μ).
From equation (2):
-0.2 = (65 - μ) / σ
Multiply both sides by σ:
-0.2σ = 65 - μ
Rearrange the equation:
μ = 65 + 0.2σ ----(3)
Now, substitute the value of μ from equation (3) into equation (1):
1.6 = (82 - (65 + 0.2σ)) / σ
Simplify:
1.6 = (82 - 65 - 0.2σ) / σ
1.6 = (17 - 0.2σ) / σ
Cross multiply:
1.6σ = 17 - 0.2σ
Multiply both sides by 10:
16σ = 170 - 2σ
Add 2σ to both sides:
16σ + 2σ = 170
Combine like terms:
18σ = 170
Divide both sides by 18:
σ ≈ 9.44
Now, substitute the value of σ back into equation (3):
μ = 65 + 0.2σ
μ = 65 + 0.2(9.44)
μ ≈ 67.89
Therefore, the mean (μ) is approximately 67.89 and the standard deviation (σ) is approximately 9.44 for the scores of all the students who took the examination.