It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minutes per day that people spend standing or walking. Among mildly obese people, the mean number of minutes of daily activity (standing or walking) is approximately Normally distributed with 376 minutes and standard deviation 70 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with 526 minutes and standard deviation 106 minutes. A researcher records the minutes of activity for an SRS of 6 mildly obese people and an SRS of 6 lean people. Required:
a. What is the probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?
b. What is the probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?

0.9963 is the probabilities that mean number of daily activity of 6 lean people exceed 410 minutes.

Step-by-step explanation:

Solution:

6 Mildly obese people exceed 410 minutes.

Given:

Mean = µ = 376

Standard deviation = ∂ = 70

n = 6

p (x> 410) = p (z > x - µ/ ∂ / √n)

                 = p (z > 410 – 376 / 70 /√6 )

                 = p (z > 1.185)

                 = 1 – p (z < 1.185)

                 = 1 – 0.38199

                = 0.618

0.618 is the probability that the mean number of minutes of daily activity of 6 mildly obese people exceed 410 minutes.

6 lean people exceed 410 minutes:

Given:

Mean = µ = 526

Standard deviation = ∂ = 106

n = 6

=p(x > 410)

= p (z > x - µ/ ∂ /√n)

=p (z > (410 – 526 ) / 106 / √ 6 )

= p (z > - 2.67)

= p (z <2.68)

= 0.9963

0.9963 is the probabilities that mean number of daily activity of 6 lean people exceed 410 minutes.

han when lo ahe whn hew anh ? ?

explanation:

it has the same slope and a different y-intercept

hope this : )