Final answer:
The provided quartiles options do not accurately reflect the normal distribution of NBA player heights. Instead, the calculation of the z-scores for given heights using the z = (X - µ) / σ formula indicates how many standard deviations a player's height is from the mean, either above or below.
Step-by-step explanation:
The question asks about labeling the normal distribution graph with values for the mean and the three quartiles (1Q, 2Q, and 3Q) for the population of NBA players with a mean height of 6'7" (79 inches) and a standard deviation of 3.9 inches. The values provided in the options (a, b, c, d) do not correspond to quartiles; instead, they seem to be just different heights. To calculate the actual quartiles, we would need to determine the points on the normal distribution that divide the population into quarters. However, since that is not provided by any of the options, the question appears to be flawed or incomplete.
As for the additional information provided on calculating z-scores, we can apply the z-score formula z = (X - µ) / σ where X is the height in inches, µ is the mean height (79 inches), and σ is the standard deviation (3.89 inches).
For a height of 77 inches, the z-score would be calculated as follows:
z = (77 - 79) / 3.89 = -0.5141.
This indicates that a height of 77 inches is 0.5141 standard deviations below the mean.
For a height of 85 inches, the z-score would be:
z = (85 - 79) / 3.89 = 1.5424.
This indicates that a height of 85 inches is 1.5424 standard deviations above the mean.
If a player reported a height with a z-score of 3.5, this would be highly unusual as it is more than three standard deviations away from the mean, and would suggest a height significantly taller than virtually all players. Such extreme heights are rare, therefore, the claim might be subject to skepticism without verification.