Mathematics, 06.04.2021 04:00 tatilynnsoto17

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 376 minutes and standard deviation 67 minutes. The mean number of minutes of daily activity for lean people is approximately Normally distributed with mean 520 minutes and standard deviation 110 minutes. A researcher records the minutes of activity for an SRS of 7 mildly obese people and an SRS of 7 lean people. A) What is the probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes?
B) What is the probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes?

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Answers

a) 0.0537 = 5.37% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes

b) 0.9871 = 98.71% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Mildly obese:

Mean 376 minutes and standard deviation 67 minutes, which means that $\mu = 376, \sigma = 67$

Sample of 6

This means that $n = 6, s = \frac{67}{\sqrt{6}} = 27.35$

Lean

Mean 520 minutes and standard deviation 110 minutes, which means that $\mu = 520, \sigma = 110$

Sample of 6

$n = 6, s = \frac{110}{\sqrt{6}} = 44.91$

A) What is the probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes?

This is 1 subtracted by the pvalue of Z when X = 420, using the mean and standard deviation for mildly obese people. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{420 - 376}{27.35}$

Z = 1.61

Z = 1.61 has a pvalue of 0.9463

1 - 0.9463 = 0.0537

0.0537 = 5.37% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 420 minutes.

B) What is the probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes?

This is 1 subtracted by the pvalue of Z when X = 420, using the mean and standard deviation for lean people. So

$Z = \frac{X - \mu}{s}$

$Z = \frac{420 - 520}{44.91}$

Z = -2.23

Z = -2.23 has a pvalue of 0.0129

1 - 0.0129 = 0.9871

0.9871 = 98.71% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 420 minutes

Answer from: Quest

1/5^2 = 7.56/45

Answer from: Quest

step-by-step explanation:

19. what two-dimensional shape is formed by a cross section of the cube shown if the cross section p

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It appears that people who are mildly obese are less active than leaner people. One study looked at...