Mathematics, 27.02.2020 22:57, BigCra

In a study on the physical activity of young adults, pediatric researchers measured overall physical activity as the total number of registered movements (counts) over a period of time and then computed the number of counts per minute (cpm) for each subject (International Journal of Obesity, January 2007). The study revealed that the overall physical activity of obese young adults is skewed right with a mean of µ = 320 cpm and a standard deviation of σ = 100 cpm. In a random sample of n = 100 obese young adults, consider the sample mean counts per minute, ¯x.

(a) Describe the sampling distribution of ¯x making sure to comment on all relevant features including its mean, standard error, and shape.
(b) What is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm?
(c) What is the probability that the mean overall physical activity level of the sample is greater than 360 cpm?

Answers: 2

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Answers

a) By the Central Limit Theorem, the sampling distribution of ¯x will be bell-shaped(which is the same as normally distributed) with mean $\mu = 320$ and standard error s = 10

b) 13.59% probability that the mean overall physical activity level of the sample is between 300 and 310 cpm

c) 0% probability that the mean overall physical activity level of the sample is greater than 360 cpm

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 normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore 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 pvalue, 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation of the sample, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 320, \sigma = 100, n = 100, s = \frac{100}{\sqrt{100}} = 10$

(a) Describe the sampling distribution of ¯x making sure to comment on all relevant features including its mean, standard error, and shape.

By the Central Limit Theorem, the sampling distribution of ¯x will be bell-shaped(which is the same as normally distributed) with mean $\mu = 320$ and standard error s = 10

(b) What is the probability that the mean overall physical activity level of the sample is between 300 and 310 cpm?

This is the pvalue of Z when X = 310 subtracted by the pvalue of Z when X = 300. So

X = 310

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

By the Central Limit Theorem

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

$Z = \frac{310 - 320}{10}$

Z = -1

Z = -1 has a pvalue of 0.1587

X = 300

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

$Z = \frac{300 - 320}{10}$

Z = -2

Z = -2 has a pvalue of 0.0228

0.1587 - 0.0228 = 0.1359

13.59% probability that the mean overall physical activity level of the sample is between 300 and 310 cpm

This is 1 subtracted by the pvalue of Z when X = 360. So

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

$Z = \frac{360 - 320}{10}$

Z = 4

Z = 4 has a pvalue of 1

1-1 = 0

0% probability that the mean overall physical activity level of the sample is greater than 360 cpm

answered: Guest

step-by-step explanation:

answered: Guest

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step-by-step explanation:

told you