Using the z-distribution, as we have a proportion, the 95% confidence interval is (0.2316, 0.3112).
<h3>What is a confidence interval of proportions?</h3>
A confidence interval of proportions is given by:
In which:
is the sample proportion.
In this problem, we have a 95% confidence level, hence
, z is the value of Z that has a p-value of 
, so the critical value is z = 1.96.
We also consider that 130 out of the 479 season ticket holders spent $1000 or more at the previous two home football games, hence:
Hence the bounds of the interval are found as follows:
The 95% confidence interval is (0.2316, 0.3112).
More can be learned about the z-distribution at brainly.com/question/25890103
Answer:
Step-by-step explanation:
4. f(x) = √(x + 4)
f(5) = √(5+4) = √(9) = 3
5. g(x) = x^12 - x^5
g(-1)= (-1)^12 - (-1)^5
1 - (-1) = 1 + 1 = 2
The answer is linear and increasing. look at it.... it goes up.
Answer:
The mean would be $322,343 and the median would be $196,723.
Step-by-step explanation:
Since the distribution of individual incomes is skewed to the right, it means that the distribution has a long right tail.
Drawing a distribution with this characteristic, we can see how the majority of the data falls into the left side of the graphic, meaning that a lot of people receive less income. Following this reasoning, the mean which is the amount of the data (in this case individual income) divided by the amount of people, would be the higher number, meaning that the few people who earn more money would influence in making this number higher.
Following this reasoning, the median (which is not influenced by this difference) would be the less high number.
