Respuesta :
Answer:
a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.
b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\frac{\sigma}{\sqrt{n}}[/tex].
Normal probability distribution
Problems of normally distributed samples can be solved using 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.
Mildly obese
Normally distributed with mean 373 minutes and standard deviation 67 minutes. So [tex]\mu = 373, \sigma = 67[/tex]
A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?
So [tex]n = 5, s = \frac{67}{\sqrt{5}} = 29.96[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 410.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{410 - 373}{29.96}[/tex]
[tex]Z = 1.23[/tex]
[tex]Z = 1.23[/tex] has a pvalue of 0.8907.
So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.
Lean
Normally distributed with mean 526 minutes and standard deviation 107 minutes. So [tex]\mu = 526, \sigma = 107[/tex]
B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?
So [tex]n = 5, s = \frac{107}{\sqrt{5}} = 47.86[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 410.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{410 - 526}{47.86}[/tex]
[tex]Z = -2.42[/tex]
[tex]Z = -2.42[/tex] has a pvalue of 0.0078.
So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.
