Answer:
(a) The mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes is 0.0495.
(b) The mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes is 0.9909.
Step-by-step explanation:
Let <em>X</em> = amount of time mildly obese people do daily activity and <em>Y</em> = amount of time lean people do daily activity.
The random variable
and
.
The sample size of mildly obese and lean people selected is, <em>n</em> = 6.
(a)
Compute the probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes as follows:
![P(\bar X->420)=P(\frac{\bar X-\mu}{\sigma/\sqrt{n}}>\frac{420-375}{67/\sqrt{6}} )\\=P(Z>1.65)\\=1-P(Z](https://tex.z-dn.net/?f=P%28%5Cbar%20X-%3E420%29%3DP%28%5Cfrac%7B%5Cbar%20X-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3E%5Cfrac%7B420-375%7D%7B67%2F%5Csqrt%7B6%7D%7D%20%29%5C%5C%3DP%28Z%3E1.65%29%5C%5C%3D1-P%28Z%3C1.65%29%5C%5C%3D1-0.95053%5C%5C%3D0.04947)
*Use the <em>z</em>-table for the probability.
Thus, the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes is 0.0495.
(b)
Compute the probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes as follows:
![P(\bar Y->420)=P(\frac{\bar Y-\mu}{\sigma/\sqrt{n}}>\frac{420-523}{107/\sqrt{6}} )\\=P(Z>-2.36)\\=P(Z](https://tex.z-dn.net/?f=P%28%5Cbar%20Y-%3E420%29%3DP%28%5Cfrac%7B%5Cbar%20Y-%5Cmu%7D%7B%5Csigma%2F%5Csqrt%7Bn%7D%7D%3E%5Cfrac%7B420-523%7D%7B107%2F%5Csqrt%7B6%7D%7D%20%29%5C%5C%3DP%28Z%3E-2.36%29%5C%5C%3DP%28Z%3C2.36%29%5C%5C%3D0.99086)
*Use the <em>z</em>-table for the probability.
Thus, the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes is 0.9909.