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).
Median = 50th percentile
To find the number of hours per week that approximately 50 percent of SOC 390 students work or more, we need to find the z-score corresponding to the percentile of 50%.
The z-score formula is given by:
z = (x - μ) / σ
Where:
x = the value we are interested in (number of hours worked per week in this case)
μ = mean of the distribution (22 in this case)
σ = standard deviation of the distribution (17 in this case)
To find the z-score corresponding to the 50th percentile, which represents the median in a normal distribution, we can use a standard normal distribution table or calculators available online. The z-score corresponding to the 50th percentile is approximately 0.
Now we can solve for x in the z-score formula:
0 = (x - 22) / 17
Rearranging the formula to solve for x, we get:
x - 22 = 0
x = 22
So, approximately 50 percent of SOC 390 students work more than or equal to 22 hours per week.