The percentage of reduction is 17%
<u>Explanation:</u>
Original price = $21
Reduced price = $17.43
Percentage of reduction = ?
Difference in price = $21 - $17.43
= $3.57
Percentage of reduction =
Therefore, percentage of reduction is 17%
Using the z-distribution, it is found that the 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).
If we had increased the confidence level, the margin of error also would have increased.
<h3>What is a confidence interval of proportions?</h3>A confidence interval of proportions is given by:
In which:
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. Increasing the confidence level, z also increases, hence the margin of error also would have increased.
The sample size and the estimate are given as follows:
.
The lower and the upper bound of the interval are given, respectively, by:
The 95% confidence interval for the proportion of college students who work to pay for tuition and living expenses is: (0.4239, 0.5161).
More can be learned about the z-distribution at brainly.com/question/25890103
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Answer:
a) P(X≥143)=0
b) This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).
Step-by-step explanation:
If we use the normal approximation to the binomial distribution we have the following parameters (mean and standard deviation):
Then, we can calculate the probability of X being equal or more than 143 using the z-score:
This contradicts the study as getting a sample with this proportion is almost impossible (if the proportion of 68% is true).