The caffeine content of a cup of home-brewed coffee is a normally distributed random variable
with a mean of 115 mg with a standard deviation of 20 mg. (a) What is the probability that a ran�domly chosen cup of home-brewed coffee will have more than 130 mg of caffeine? (b) Less than
100 mg? (c) A very strong cup of tea has a caffeine content of 91 mg. What is the probability that
a cup of coffee will have less caffeine than a very strong cup of tea?
To answer these questions, we need to use the concept of the z-score and the standard normal distribution. The z-score measures how many standard deviations an observation is from the mean.
Given the mean (μ) of 115 mg and the standard deviation (σ) of 20 mg, we can calculate the z-score for each of the values of interest using the formula:
z = (x - μ) / σ
where x is the value we are interested in.
(a) To find the probability that a randomly chosen cup of home-brewed coffee will have more than 130 mg of caffeine, we need to find the area under the normal curve to the right of 130 mg.
z = (130 - 115) / 20 = 0.75
We can use a standard normal distribution table or software to find that the probability corresponding to a z-score of 0.75 is approximately 0.2266. Therefore, the probability that a randomly chosen cup of home-brewed coffee will have more than 130 mg of caffeine is 0.2266 or 22.66%.
(b) To find the probability that a randomly chosen cup of home-brewed coffee will have less than 100 mg of caffeine, we need to find the area under the normal curve to the left of 100 mg.
z = (100 - 115) / 20 = -0.75
Again, using the standard normal distribution table or software, we can find that the probability corresponding to a z-score of -0.75 is approximately 0.2266. However, since we are interested in the probability of less than 100 mg (to the left of 100 mg), we need to subtract this probability from 1.
Probability = 1 - 0.2266 = 0.7734 or 77.34%
Therefore, the probability that a randomly chosen cup of home-brewed coffee will have less than 100 mg of caffeine is 0.7734 or 77.34%.
(c) To find the probability that a cup of coffee will have less caffeine than a very strong cup of tea (91 mg), we need to find the area under the normal curve to the left of 91 mg.
z = (91 - 115) / 20 = -1.2
Using the standard normal distribution table or software, we can find that the probability corresponding to a z-score of -1.2 is approximately 0.1151. Therefore, the probability that a randomly chosen cup of home-brewed coffee will have less caffeine than a very strong cup of tea (91 mg) is 0.1151 or 11.51%.
I hope this explanation helps you understand how to calculate these probabilities using the z-score and the standard normal distribution.