Question

The average daily caffeine consumption for an adult follows an approximately normal distribution with mean of 135 milligrams and standard deviation of 24 milligrams. Show work or include a calculator or Excel command to receive credit for each part.

a. What percentage of people consume an average less than 148.2 milligrams of caffeine?
b. What is probability a person consumes an average between 99 and 178.2 milligrams of caffeine?
c. What average consumption represents the 88.88 th percentile?
d. What average consumption levels represents the middle 92% of consumption levels?

Answer 1

a. The percentage of people who consume an average less than 148.2 milligrams of caffeine is approximately 70.00%. b. The probability a person consumes an average between 99 and 178.2 milligrams of caffeine is approximately 0.9032. c. The average consumption that represents the 88.88th percentile is approximately 161.32 milligrams. d. The average consumption levels that represent the middle 92% of consumption levels are approximately 95.5 and 174.5 milligrams.

Step-by-step explanation:

a. To find the percentage of people who consume an average less than 148.2 milligrams of caffeine, we need to calculate the z-score using the formula: z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation. Substituting the values, we get z = (148.2 - 135) / 24 = 0.55. Using a standard normal table or calculator, we can find that the percentage is approximately 70.00%.

b. To find the probability a person consumes an average between 99 and 178.2 milligrams of caffeine, we need to calculate the z-scores for both values. The z-score for 99 is z = (99 - 135) / 24 = -1.50, and the z-score for 178.2 is z = (178.2 - 135) / 24 = 1.80. Using the standard normal table or calculator, we can find the area to the left of each z-score and subtract the smaller value from the larger value to find the probability, which is approximately 0.9032.

c. To find the average consumption that represents the 88.88th percentile, we need to find the z-score that corresponds to this percentile using the standard normal table or calculator. The z-score is approximately 1.18. Using the z-score formula, we can solve for x: 1.18 = (x - 135) / 24. Solving for x, we get x = 135 + (1.18 * 24) = 161.32 milligrams.

d. To find the average consumption levels that represent the middle 92% of consumption levels, we need to find the z-scores that correspond to the lower and upper percentiles of 4% (100% - 92%). Using the standard normal table or calculator, we find that the z-score for the lower percentile is approximately -1.75 and the z-score for the upper percentile is approximately 1.75. Using the z-score formula, we can solve for x: -1.75 = (x - 135) / 24 and 1.75 = (x - 135) / 24. Solving for x in each equation, we get x = 95.5 and x = 174.5 milligrams, respectively.