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 randomly 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 find the probabilities in this scenario, we can use the standard normal distribution and the z-score formula. The z-score formula is given by:
z = (x - μ) / σ
Where:
- z is the z-score,
- x is the value we are interested in,
- μ is the mean of the distribution,
- σ is the standard deviation of the distribution.
(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 compute the area under the curve to the right of the z-score associated with 130 mg.
First, calculate the z-score of 130 mg:
z = (130 - 115) / 20
z = 0.75
Next, we want to find the probability to the right of z = 0.75 on the standard normal distribution table. This probability corresponds to the area under the curve beyond z = 0.75.
Using a standard normal distribution table or calculator, we find that the probability associated with z = 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 compute the area under the curve to the left of the z-score associated with 100 mg.
First, calculate the z-score of 100 mg:
z = (100 - 115) / 20
z = -0.75
Next, we want to find the probability to the left of z = -0.75 on the standard normal distribution table. This probability corresponds to the area under the curve to the left of z = -0.75.
Using a standard normal distribution table or calculator, we find that the probability associated with z = -0.75 is approximately 0.2266.
Therefore, the probability that a randomly chosen cup of home-brewed coffee will have less than 100 mg of caffeine is 0.2266, or 22.66%.
(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 compute the area under the curve to the left of the z-score associated with 91 mg.
First, calculate the z-score of 91 mg:
z = (91 - 115) / 20
z = -1.2
Next, we want to find the probability to the left of z = -1.2 on the standard normal distribution table. This probability corresponds to the area under the curve to the left of z = -1.2.
Using a standard normal distribution table or calculator, we find that the probability associated with z = -1.2 is approximately 0.1151.
Therefore, the probability that a cup of coffee will have less caffeine than a very strong cup of tea is 0.1151, or 11.51%.