To decide the mean of the test, we can utilize the data given about the standard deviations and the scores.
How about we indicate the mean as "μ," the standard deviation as "σ," and the given scores as follows:
A score of 52 falls three standard deviations below the mean, which can be expressed as
52 = μ - 3σ
A score of 62 falls one standard deviation below the mean, which can be expressed as:
62 = μ - σ
Now, we have a system of two equations:
Equation 1: 52 = μ - 3σ
Equation 2: 62 = μ - σ
We can use Equation 2 to express μ in terms of σ:
μ = 62 + σ
Now, we can substitute this expression for μ into Equation 1:
52 = (62 + σ) - 3σ
Now, solve for σ:
52 = 62 + σ - 3σ
Combine like terms:
52 = 62 - 2σ
Subtract 62 from both sides:
-10 = -2σ
Now, divide by -2 to solve for σ:
σ = 5