A chameleon is looking for prey. Let positive numbers represent the elevation of prey above the chameleon and negative numbers r
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Answer:
The area of the sector is
Step-by-step explanation:
we know that
The area of a complete circle (16π units^2) subtends a central angle of 2π radians
so
using proportion
Find out the area of a sector , if the central angle is equal to 8π/5 radians
Answer:
10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.
99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
Mildly obese
Normally distributed with mean 375 minutes and standard deviation 68 minutes. So
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.8962.
So there is a 1-0.8962 = 0.1038 = 10.38% probability that the mean number of minutes of daily activity of the 6 mildly obese people exceeds 410 minutes.
Lean
Normally distributed with mean 522 minutes and standard deviation 106 minutes. So
What is the probability (±0.0001) that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.0045.
So there is a 1-0.0045 = 0.9955 = 99.55% probability that the mean number of minutes of daily activity of the 6 lean people exceeds 410 minutes
You then divide the numerator(top) by the denominator(bottom)
Your answer is