A population of values has a normal distribution with μ=153.9μ
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A population of values has a
A) Find P86, which is the score separating the bottom 86% scores from the top 14% scores.
P86 (for single values) =
B) Find P86, which is the mean separating the bottom 86% means from the top 14% means.
P86 (for sample means) =
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- A population of values has a normal distribution with μ=128 and σ=32.4 You intend to draw a random sample of size n=120.Find P45, which is the mean separating the bottom 45% means from the top 55% means.P45 (for sample means) =A population of values has a normal distribution with μ=248.4μ=248.4 and σ=23.6σ=23.6. You intend to draw a random sample of size n=32n=32.Find P18, which is the score separating the bottom 18% scores from the top 82% scores.P18 (for single values) = Find P18, which is the mean separating the bottom 18% means from the top 82% means.P18 (for sample means) = Enter your answers as numbers accurate to 1 decimal place.************NOTE************ round your answer to ONE digit after the decimal point! ***********Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.A population of values has a normal distribution with μ=106.2μ=106.2 and σ=61σ=61. You intend to draw a random sample of size n=85n=85.Find P93, which is the score separating the bottom 93% scores from the top 7% scores.P93 (for single values) = Find P93, which is the mean separating the bottom 93% means from the top 7% means.P93 (for sample means) = Enter your answers as numbers accurate to 1 decimal place.
- A population of values has a normal distribution with μ=158.8μ=158.8 and σ=91.9σ=91.9. You intend to draw a random sample of size n=141n=141.Inverse Normal CalculatorFind P10, which is the score separating the bottom 10% scores from the top 90% scores.P10 (for single values) = Find P10, which is the mean separating the bottom 10% means from the top 90% means.P10 (for sample means) = Enter your answers as numbers accurate to 1 decimal place.A population of values has a normal distribution with μ=185.7μ=185.7 and σ=57.5σ=57.5. You intend to draw a random sample of size n=57n=57.Find P25, which is the score separating the bottom 25% scores from the top 75% scores.P25 (for single values) = Find P25, which is the mean separating the bottom 25% means from the top 75% means.P25 (for sample means) =A population of values has a normal distribution with μ=123.9μ=123.9 and σ=17.2σ=17.2. You intend to draw a random sample of size n=17n=17.Find P85, which is the score separating the bottom 85% scores from the top 15% scores.P85 (for single values) = Find P85, which is the mean separating the bottom 85% means from the top 15% means.P85 (for sample means) = Enter your answers as numbers accurate to 1 decimal place.************NOTE************ round your answer to ONE digit after the decimal point! ***********Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.
- For a population that has a standard deviation of 22, figure the standard deviation of the distribution of means for samples of size (a)2, (b)4, (c)5 and (d)10.Justin wants to know whether a commonly prescribed drug does improve the attention span of students with attention deficit disorder (ADD). He knows that the mean attention span for students with ADD who are not taking the drug is 2.3 minutes long. His sample of 12 students taking the drug yielded a mean of 4.6 minutes. Justin can find no information regarding σx , so he calculated s2x =1.96. a. Identify the independent and dependent variables. b. In a sentence, state the null hypothesis being tested. c. Using symbols, state the null and alternative hypotheses. d. Determine the critical region using a one-tailed test with alpha = .05. e .Conduct the hypothesis test (Do the math and compare the t-critical and t-obtained values). f. State your conclusions in terms of H0 (Should you reject the H0 or fail to reject/accept the H0). g. Based on your analysis, is there a relationship between the drug and attention span?Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 24 samples is 8.1 ppm with a variance of 0.25. Does the data support the claim at the 0.1 level? Assume the population distribution is approximately normal. Step 5 of 5: Make the decision to reject or fail to reject the null hypothesis.
