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 and standard deviation
, the z-score of a measure X is given by:
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 and standard deviation
, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
.
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
Sample of 6
This means that
Lean
Mean 520 minutes and standard deviation 110 minutes, which means that
Sample of 6
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
By the Central Limit Theorem
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
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
