First, you subtract 2x on both sides
divided on both sides by -3
divided on both sides by -3
This scatter plot show the relationship between the temperature and amount of sales at a store. The line of best fit is shown on
the graph.
a. The y-intercept of the estimated line of best fit is at (0,b). Enter the approximate value of b. Round your estimate to the nearest whole number.
b. Enter the approximate slope of the estimated line of best fit in the second box.
a. The y-intercept of the estimated line of best fit is at (0,b). Enter the approximate value of b. Round your estimate to the nearest whole number.
b. Enter the approximate slope of the estimated line of best fit in the second box.
1 answer:
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Answer:
a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.
b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 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 373 minutes and standard deviation 67 minutes. So
A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.8907.
So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.
Lean
Normally distributed with mean 526 minutes and standard deviation 107 minutes. So
B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?
So
This probability is 1 subtracted by the pvalue of Z when X = 410.
has a pvalue of 0.0078.
So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.