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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
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