In a population of exam scores, a score of X=88 corresponds to z= +2.00 and a score of X=79 corresponds to z= -1.00. Find the mean and standard deviation for the population.
To find the mean and standard deviation for the population, we need to use the formula:
z = (X - μ) / σ
where:
- z is the z-score,
- X is the raw score,
- μ is the population mean, and
- σ is the population standard deviation.
We have two data points to work with:
1. When X = 88, z = +2.00.
2. When X = 79, z = -1.00.
Using the given information, we can set up two equations:
Equation 1: 2.00 = (88 - μ) / σ (1)
Equation 2: -1.00 = (79 - μ) / σ (2)
We can now solve these two equations simultaneously.
From Equation (1):
2.00 = (88 - μ) / σ
2.00σ = 88 - μ (3)
From Equation (2):
-1.00 = (79 - μ) / σ
-σ = 79 - μ (4)
Solving Equations (3) and (4) together, we get:
2.00σ = 88 - μ (3)
-σ = 79 - μ (4)
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1.00σ = 9 - μ (5)
Rearranging Equation (5) gives us:
μ = 9 - 1.00σ (6)
Substituting this value of μ into Equation (3), we have:
2.00σ = 88 - (9 - 1.00σ)
2.00σ = 88 - 9 + 1.00σ
1.00σ = 79
Therefore, σ = 79.
Substituting this value of σ into Equation (6), we find:
μ = 9 - 1.00(79)
μ = 9 - 79
μ = -70
Therefore, the population mean (μ) is -70, and the population standard deviation (σ) is 79.
To find the mean and standard deviation of a population, we can use the z-score formula:
z = (X - μ) / σ
Where:
- X is the raw score
- μ is the mean of the population
- σ is the standard deviation of the population
We are given two data points with their corresponding z-scores:
For X = 88, z = +2.00
For X = 79, z = -1.00
Using the given z-score formula, we can set up two equations:
2.00 = (88 - μ) / σ
-1.00 = (79 - μ) / σ
Now we can solve this system of equations to find the values of μ (mean) and σ (standard deviation).
Let's start by rearranging the first equation:
2.00σ = 88 - μ
Next, rearrange the second equation:
-1.00σ = 79 - μ
Now we have a system of linear equations:
2.00σ = 88 - μ
-1.00σ = 79 - μ
To eliminate μ, we can multiply the second equation by 2:
2*(-1.00σ) = 2*(79 - μ)
-2.00σ = 158 - 2μ
Now we can subtract the two equations:
2.00σ - (-2.00σ) = 88 - (-2μ) - (158 - 2μ)
4.00σ = 246 - 2μ + 2μ
4.00σ = 246
Now isolate σ by dividing both sides by 4.00:
σ = 246 / 4.00
σ = 61.5
So, the standard deviation of the population is 61.5.
To find the mean (μ), substitute the value of σ into one of the equations (let's use the first one):
2.00 = (88 - μ) / 61.5
Now, multiply both sides by 61.5:
2.00 * 61.5 = 88 - μ
122.9 = 88 - μ
Now isolate μ by subtracting 88 from both sides:
122.9 - 88 = - μ
34.9 = - μ
To remove the negative sign, multiply both sides by -1:
34.9 * (-1) = (- μ) * (-1)
-34.9 = μ
So, the mean of the population is -34.9.
Therefore, the mean of the population is -34.9 and the standard deviation is 61.5.