Answer:
10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.
99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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 375 minutes and standard deviation 68 minutes. So ![\mu = 375, \sigma = 68](https://tex.z-dn.net/?f=%5Cmu%20%3D%20375%2C%20%5Csigma%20%3D%2068)
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?
So ![n = 6, s = \frac{68}{\sqrt{6}} = 27.76](https://tex.z-dn.net/?f=n%20%3D%206%2C%20s%20%3D%20%5Cfrac%7B68%7D%7B%5Csqrt%7B6%7D%7D%20%3D%2027.76)
This probability is 1 subtracted by the pvalue of Z when X = 410.
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{410 - 375}{27.76}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B410%20-%20375%7D%7B27.76%7D)
![Z = 1.26](https://tex.z-dn.net/?f=Z%20%3D%201.26)
has a pvalue of 0.8962.
So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.
Lean
Normally distributed with mean 522 minutes and standard deviation 106 minutes. So ![\mu = 522, \sigma = 106](https://tex.z-dn.net/?f=%5Cmu%20%3D%20522%2C%20%5Csigma%20%3D%20106)
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?
So ![n = 6, s = \frac{106}{\sqrt{6}} = 43.27](https://tex.z-dn.net/?f=n%20%3D%206%2C%20s%20%3D%20%5Cfrac%7B106%7D%7B%5Csqrt%7B6%7D%7D%20%3D%2043.27)
This probability is 1 subtracted by the pvalue of Z when X = 410.
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{410 - 523}{43.27}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B410%20-%20523%7D%7B43.27%7D)
![Z = -2.61](https://tex.z-dn.net/?f=Z%20%3D%20-2.61)
has a pvalue of 0.0045.
So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes