Question
1.) According to the American College Test (ACT), results from the 2004 ACT testing found that students had a mean reading score of 21.3 and a standard deviation of 6.0. The scores are normally distributed. A student scores a 35. What percent of the data would contain that score?
2.) Find the margin of error for the sample. Then find the interval that would likely contain the true population proportion.
47% of 296 teens
3.) A and B are events defined on a sample space, with P(A) = 0.6 and P(A and B) = 0.3. Find the P(B|A).
4.) Find the margin of error for the sample. Then find the interval that would likely contain the true population proportion.
23% of 100 bakers
5.) Juan lives in Washington DC. He travels by speed rail 80% of the time. Other days he takes a taxi 20% of the time even though it is more expensive. When taking the speed rail, he arrives at work on time 70% of the time. When taking the taxi, he arrives on time 90% of the time. Whether Juan takes the rail or taxi, what is the probability he arrives on time for work on any given day?
Thanks!!! <3
2.) Find the margin of error for the sample. Then find the interval that would likely contain the true population proportion.
47% of 296 teens
3.) A and B are events defined on a sample space, with P(A) = 0.6 and P(A and B) = 0.3. Find the P(B|A).
4.) Find the margin of error for the sample. Then find the interval that would likely contain the true population proportion.
23% of 100 bakers
5.) Juan lives in Washington DC. He travels by speed rail 80% of the time. Other days he takes a taxi 20% of the time even though it is more expensive. When taking the speed rail, he arrives at work on time 70% of the time. When taking the taxi, he arrives on time 90% of the time. Whether Juan takes the rail or taxi, what is the probability he arrives on time for work on any given day?
Thanks!!! <3
Answers
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
However, I will start you out with the first problem. I assume that you want the proportion below.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability below that Z score.
However, I will start you out with the first problem. I assume that you want the proportion below.
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability below that Z score.
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