Question

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.10 Among mildly obese people, minutes of activity varied according to the N(373, 67) distribution. Minutes of activity for lean people had the N(526, 107) distribution. Within what limits do the active minutes for about 95% of the people in each group fall? Use the 68–95–99.7 rule.

Answer 1

Obese people

`Lower = \mu - 2\sigma = 373- 2(67) = 239`

`Upper = \mu + 2\sigma = 373+ 2(67) = 507`

Lean People

`Lower = \mu - 2\sigma = 526- 2(107) = 312`

`Upper = \mu + 2\sigma = 526+ 2(107) = 740`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, "almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ)".

Solution to the problem

Obese people

Let X the random variable that represent the minutes of a population (obese people), and for this case we know the distribution for X is given by:

`X \sim N(373,67)`

Where
`\mu=373` and
`\sigma=67`

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

`Lower = \mu - 2\sigma = 373- 2(67) = 239`

`Upper = \mu + 2\sigma = 373+ 2(67) = 507`

Lean People

Let X the random variable that represent the minutes of a population (lean people), and for this case we know the distribution for X is given by:

`X \sim N(526,107)`

Where
`\mu=526` and
`\sigma=107`

On this case we know that 95% of the data values are within two deviation from the mean using the 68-95-99.7 rule so then we can find the limits liek this:

`Lower = \mu - 2\sigma = 526- 2(107) = 312`

`Upper = \mu + 2\sigma = 526+ 2(107) = 740`

The interval for the lean people is significantly higher than the interval for the obese people.