If a random sample of 20 persons weighed 3,460, the sample mean x-bar would be 3460/20 = 173 pounds.
The z-score for 173 pounds is given by:
Referring to a standard normal distribution table, and using z = 0.66, we find:
Therefore
The answer is: 0.2546
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
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Answer:
the answer is 4
Step-by-step explanation:
24* 80/100 = 24*8/10 = 19.2 is the answer.