Let's pretend the Work variable is normally distributed, with a mean/median/mode value of 22 and a standard deviation of 17. Based on these numbers, we can say that approximately 50 percent of SOC 390 students work more than ____ hours per week (round answer to the nearest WHOLE number, no decimals).
To determine the number of hours per week that approximately 50 percent of SOC 390 students work more than, we need to use the concept of the standard normal distribution.
The standard normal distribution has a mean of 0 and a standard deviation of 1. In order to work with the data given, we need to convert it to the standard normal distribution.
We can do this by using the formula for z-score:
z = (x - μ) / σ
where:
z = the z-score
x = the observed value
μ = the mean
σ = the standard deviation
In this case, we are looking for the z-score that corresponds to the value where 50 percent of students work more than. Since we know that the mean/median/mode value is 22, we can use this value for x.
z = (x - μ) / σ
z = (22 - 22) / 17
z = 0 / 17
z = 0
From the standard normal distribution table, we know that at a z-score of 0, the cumulative probability is 0.5, which means 50 percent of the data falls below this value.
Now, we need to convert this z-score back to the original scale of the Work variable. To do this, we can use the formula:
x = μ + (z * σ)
Substituting the values:
x = 22 + (0 * 17)
x = 22
Therefore, approximately 50 percent (half) of SOC 390 students work more than 22 hours per week.