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
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.
1.) To find the percent of data that would contain a score of 35, we need to calculate the z-score for that score and then find the corresponding percentile.
First, we calculate the z-score using the formula:
z = (x - μ) / σ
where x is the score, μ is the mean, and σ is the standard deviation.
z = (35 - 21.3) / 6.0
z ≈ 2.2833
Using a z-table or a statistical calculator, we can find that the percentile corresponding to a z-score of 2.2833 is approximately 98.9%. This means that approximately 98.9% of the data would have scores less than or equal to 35.
2.) To find the margin of error for a sample, we need to use the formula:
margin of error = z * √(p̂ * (1 - p̂) / n)
where z is the z-score corresponding to the desired confidence level (e.g., 95% confidence corresponds to a z-score of approximately 1.96), p̂ is the sample proportion, and n is the sample size.
In this case, the sample proportion is 47% (or 0.47) and the sample size is 296. Let's assume we want a 95% confidence level (which corresponds to a z-score of approximately 1.96).
Now we can calculate the margin of error:
margin of error = 1.96 * √(0.47 * (1 - 0.47) / 296)
margin of error ≈ 0.0552
To find the interval that would likely contain the true population proportion, we can use the formula:
interval = p̂ ± margin of error
interval ≈ 0.47 ± 0.0552
So, the interval would likely contain the true population proportion of teenagers between approximately 0.4148 (0.47 - 0.0552) and 0.5252 (0.47 + 0.0552).
3.) To find the conditional probability P(B|A), we can use the formula:
P(B|A) = P(A and B) / P(A)
In this case, P(A) is given as 0.6 and P(A and B) is given as 0.3.
P(B|A) = 0.3 / 0.6
P(B|A) = 0.5
So, the probability of event B given event A is 0.5.
4.) Similar to the second question, we need to find the margin of error for the sample proportion. We can use the same formula as before:
margin of error = z * √(p̂ * (1 - p̂) / n)
Here, the sample proportion is 23% (or 0.23) and the sample size is 100. Assuming a 95% confidence level, we use the z-score of approximately 1.96:
margin of error = 1.96 * √(0.23 * (1 - 0.23) / 100)
margin of error ≈ 0.0429
The interval would likely contain the true population proportion of bakers between approximately 0.1871 (0.23 - 0.0429) and 0.2729 (0.23 + 0.0429).
5.) To calculate the probability that Juan arrives on time for work on any given day, we need to consider the probabilities of taking the speed rail and the taxi, as well as the probabilities of arriving on time given each mode of transportation.
Let's denote the probability of taking the speed rail as P(SR) = 0.8 and the probability of taking the taxi as P(Taxi) = 0.2.
The probability of arriving on time when taking the speed rail is P(On time | SR) = 0.7, and the probability of arriving on time when taking the taxi is P(On time | Taxi) = 0.9.
Using the law of total probability, we can calculate the overall probability of arriving on time:
P(On time) = P(On time | SR) * P(SR) + P(On time | Taxi) * P(Taxi)
P(On time) = 0.7 * 0.8 + 0.9 * 0.2
P(On time) = 0.56 + 0.18
P(On time) = 0.74
So, the probability that Juan arrives on time for work on any given day is 0.74 or 74%.