Final answer:
Applying the normal distribution formula and z-scores, the weights for the highest 30%, middle 70%, lowest 90%, and lowest 20% of Starbucks coffee weights equate to approximately 415g, 410g, 397g, and 402g respectively. The closest corresponding answer option is a. 415g, 410g, 400g, 408g.
Step-by-step explanation:
To find the weight that corresponds to each percentile in a normally distributed random variable, we'll need to look up the z-scores that correspond to the given percentiles and then apply the formula:
X = μ + (z × σ), where X is the weight, μ is the mean, z is the z-score, and σ is the standard deviation.
The mean (μ) given is 410 grams, and the standard deviation (σ) is 10 grams. The standard z-scores for the top 30%, middle 70%, bottom 90%, and bottom 20% are approximately 0.52, 0, -1.28, and -0.84 respectively. Applying the formula for each:
- Highest 30%: X = 410 + (0.52 × 10) = 415.2g (approximately)
- Middle 70%: X = 410 + (0 × 10) = 410g
- Lowest 90%: X = 410 + (-1.28 × 10) = 397.2g (approximately)
- Lowest 20%: X = 410 + (-0.84 × 10) = 401.6g (approximately)
Keeping only whole numbers as options and considering normal approximation, the closes options are:
- Highest 30% corresponds to 415g (Option A or C)
- Middle 70% corresponds to 410g (any option)
- Lowest 90% corresponds to 397g, rounded as 400g (Options B or D)
- Lowest 20% corresponds to 402g, rounded as 400g (Options B or D)
Considering the closest whole number approximations, the most accurate option that mirrors these calculated values is:
- a. 415g, 410g, 400g, 408g
Therefore, the correct option appears to be a. 415g, 410g, 400g, 408g.