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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 mean 371 minutes and standard deviation 65 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with mean 528 minutes and standard deviation 108 minutes. A researcher records the minutes of activity for an SRS of 6 mildly obese people and an SRS of 6 lean people.Use z-scores rounded to two decimal places to answer the following:What is the probability (Image for It appears that people who are mildly obese are less active than leaner people. One study looked at the averag0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes? What is the probability (Image for It appears that people who are mildly obese are less active than leaner people. One study looked at the averag0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes?

Respuesta :

Answer:

0.0322; 0.9929

Step-by-step explanation:

Since the data is normally distributed, we use z scores for these probabilities.

The formula for a z score of a sample mean is

[tex]z=\frac{\bar{X}-\mu}{\sigma \div \sqrt{n}}[/tex]

For the sample of mildly obese people, the mean, μ, is 371; the standard deviation, σ, is 65; and the sample size, n, is 6.

Using 420 for X,

z = (420-371)/(65÷√6) = 49/(65÷2.4495) = 49/26.5360 ≈ 1.85

Using a z table, we see that the area under the curve to the left of this is 0.9678.  However, we want the area to the right, so we subtract from 1:

1-0.9678 = 0.0322

For the sample of lean people, the mean, μ, is 528; the standard deviation, σ, is 108; the sample size, n, is 6.

Using 420 for X, we have

z = (420-528)/(108÷√6) = -108/(108÷2.4495) = -108/44.0906 ≈ -2.45

Using a z table, we see that the area under the curve to the left of this is 0.0071.  We want the area under the curve to the right, so we subtract from 1:

1-0.0071 = 0.9929

Using the normal probability distribution and the central limit theorem, it is found that there is a:

  • 0.0322 = 3.22% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes.
  • 0.9929 = 99.29% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, for the sampling distribution of sample means of size n, the standard deviation is [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]

For mildly obese people:

  • Mean of 371 minutes, thus [tex]\mu = 371[/tex].
  • Standard deviation of 108 minutes, thus [tex]\sigma = 65[/tex].
  • SRS of 6, thus, [tex]n = 6, s = \frac{65}{\sqrt{6}}[/tex]

The probability is 1 subtracted by the p-value of Z when X = 420, thus:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{420 - 371}{\frac{65}{\sqrt{6}}}[/tex]

[tex]Z = 1.85[/tex]

[tex]Z = 1.85[/tex] has a p-value of 0.9678.

1 - 0.9678 = 0.0322

0.0322 = 3.22% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes.

For lean people:

  • Mean of 528 minutes, thus [tex]\mu = 528[/tex].
  • Standard deviation of 108 minutes, thus [tex]\sigma = 108[/tex].
  • SRS of 6, thus, [tex]n = 6, s = \frac{108}{\sqrt{6}}[/tex]

The probability is 1 subtracted by the p-value of Z when X = 420, thus:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

By the Central Limit Theorem

[tex]Z = \frac{X - \mu}{s}[/tex]

[tex]Z = \frac{420 - 528}{\frac{108}{\sqrt{6}}}[/tex]

[tex]Z = -2.45[/tex]

[tex]Z = -2.45[/tex] has a p-value of 0.0071.

1 - 0.0071 = 0.9929.

0.9929 = 99.29% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes.

A similar problem is https://brainly.com/question/13499934

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