Final answer:
The weight at the 33rd percentile for a sample of male students with an average weight of 155 pounds and a standard deviation of 15 pounds is approximately 148.4 pounds. This is found using the z-score for the 33rd percentile and the given average and standard deviation.
Step-by-step explanation:
The student's question is about finding the 33rd percentile weight for a sample of male students given the average weight is 155 pounds, and the standard deviation is 15 pounds. To find the 33rd percentile, we need to use the z-score corresponding to the 33rd percentile in the standard normal distribution, then translate that into the weight using the sample's average and standard deviation.
A z-score table or calculator can tell us the z-score for the 33rd percentile. We know that the z-score is the number of standard deviations away from the mean an observation is. For the 33rd percentile, the z-score is approximately -0.44. We then use the following formula to calculate the weight at that percentile:
Z = (X - μ) / σ
Where:
Z is the z-score (-0.44 for the 33rd percentile)
μ is the mean (155 pounds in this case)
σ is the standard deviation (15 pounds here)
X is the unknown weight at the 33rd percentile
Substitute our values into the formula and solve for X:
-0.44 = (X - 155) / 15
Multiplying both sides by 15 gives us:
-6.6 = X - 155
Adding 155 to both sides we find:
X = 148.4
Thus, the weight at the 33rd percentile is approximately 148.4 pounds.