Data Set 7-5: Suppose you knew that a population had a mean of 150 and a standard deviation of 50.
-The proportion of the population of scores in Data Set 7-5 that is between 100 and 175 would be
A) .3413
B) .1915
C) .1498
D) .4706

Question

Quizplus · Accepted Answer

To find the proportion of the population with scores between 100 and 175, we use the standard deviation and mean to convert these scores into z-scores. For 100: $z = \frac{100 - 150}{50} = -1$, and for 175: $z = \frac{175 - 150}{50} = 0.5$. Looking up these z-scores in a standard normal distribution table, the area between z = -1 and z = 0 is 0.3413, and the area between z = 0 and z = 0.5 is 0.1915. Adding these areas together gives 0.3413 + 0.1915 = 0.5328. However, the closest option provided is D) .4706, which suggests a calculation or interpretation error in the explanation. The correct approach involves finding the area under the standard normal curve between the z-scores of -1 and 0.5. The correct calculation directly finds this area or closely approximates it, which is not accurately reflected in the provided options. The explanation aimed to demonstrate the process of converting raw scores to z-scores and summing areas under the normal curve, but the precise answer should align with standard normal distribution properties, indicating a need for correction in the explanation or reevaluation of the options based on accurate z-score areas.

Data Set 7-5: Suppose You Knew That a Population Had