Respuesta :
Answer:
a) 0.9332 = 93.32% drink 2 cups per day or more.
b) 0.8413 = 84.13% drink no more than 4 cups per day
c) The minimum number of cups consumed by a heavy coffee drinker is 4.52.
d) 86.86% probability that the mean number of cups per day is greater than 3
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question, we have that:
[tex]\mu = 3.2, \sigma = 0.8[/tex]
a. What proportion drink 2 cups per day or more?
This is 1 subtracted by the pvalue of Z when X = 2. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{2 - 3.2}{0.8}[/tex]
[tex]Z = -1.5[/tex]
[tex]Z = -1.5[/tex] has a pvalue of 0.0668
1 - 0.0668 = 0.9332
0.9332 = 93.32% drink 2 cups per day or more.
b. What proportion drink no more than 4 cups per day?
This is the pvalue of Z when X = 4.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{4 - 3.2}{0.8}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
0.8413 = 84.13% drink no more than 4 cups per day
c. If the top 5% of coffee drinkers are considered "heavy" coffee drinkers, what is the minimum number of cups consumed by a heavy coffee drinker?
This is the 100 - 5 = 95th percentile, which is X when Z has a pvalue of 0.95. So X when Z = 1.645. Then
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 3.2}{0.8}[/tex]
[tex]X - 3.2 = 1.645*0.8[/tex]
[tex]X = 4.52[/tex]
The minimum number of cups consumed by a heavy coffee drinker is 4.52.
d. If a sample of 20 men is selected, what is the probability that the mean number of cups per day is greater than 3?
Sample of 20, so applying the central limit theore with n = 20, [tex]s = \frac{0.8}{\sqrt{20}} = 0.1789[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 3.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3 - 3.2}{0.1789}[/tex]
[tex]Z = -1.12[/tex]
[tex]Z = -1.12[/tex] has a pvalue of 0.1314
1 - 0.1314 = 0.8686
86.86% probability that the mean number of cups per day is greater than 3
