Welcome to Westonci.ca, where you can find answers to all your questions from a community of experienced professionals. Experience the ease of finding accurate answers to your questions from a knowledgeable community of professionals. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Answer:
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 [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]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
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 [tex]\mu = 376, \sigma = 67[/tex]
Sample of 6
This means that [tex]n = 6, s = \frac{67}{\sqrt{6}} = 27.35[/tex]
Lean
Mean 520 minutes and standard deviation 110 minutes, which means that [tex]\mu = 520, \sigma = 110[/tex]
Sample of 6
[tex]n = 6, s = \frac{110}{\sqrt{6}} = 44.91[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{420 - 376}{27.35}[/tex]
[tex]Z = 1.61[/tex]
[tex]Z = 1.61[/tex] 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
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{420 - 520}{44.91}[/tex]
[tex]Z = -2.23[/tex]
[tex]Z = -2.23[/tex] 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
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Get the answers you need at Westonci.ca. Stay informed with our latest expert advice.
