Percentage grades in a large physics class follow a normal distribution with mean 58. It is known that 3% of students receive grades less than 39.2. What is the standard deviation of grades in the class?
You have to find the z-score for .03 from a table. Substitute it below to solve for standard dev.
z-score = (39.2 - 58)/stand.dev.
To find the standard deviation of the grades in the class, we can use the properties of the normal distribution and the given information.
Step 1: Convert the given percentage to a z-score.
We know that 3% of students receive grades less than 39.2. To find the corresponding z-score, we can use a standard normal distribution table or a calculator.
The z-score formula is:
z = (x - μ) / σ,
where z is the z-score, x is the value in question, μ is the mean, and σ is the standard deviation.
In this case, we can set up the equation as follows:
z = (39.2 - 58) / σ,
where z = -2.33 (since it corresponds to the lower 3% in the distribution, which is considered the left tail).
Step 2: Find the z-score that corresponds to a cumulative probability of 97%.
We subtract -2.33 from 0 (the mean) to find the z-score for a cumulative probability of 97%.
Step 3: Use the properties of the normal distribution to find the standard deviation.
From the z-score table or calculator, we find that the z-score corresponding to a cumulative probability of 97% is approximately 1.88.
We know that:
z = (x - μ) / σ.
Substituting the values we have:
1.88 = (0 - 58) / σ.
Solving for σ, we multiply both sides by σ:
1.88σ = -58.
Finally, divide both sides by 1.88:
σ ≈ -58 / 1.88.
The approximate standard deviation of grades in the class is σ ≈ -30.85.
Note: In instances like this, the negative sign on the standard deviation is typically ignored, as it doesn't affect the magnitude or interpretation of the value. The standard deviation is always positive. Thus, we can round the standard deviation to σ ≈ 30.85.