For 800 respondents to an inquiry of their age, results show a mean of 42 years...
For 800 respondents to an inquiry of their age, results show a mean of 42 years and a standard deviation of 5 years. How many respondents fall between +/- 2 standard deviation from the mean? Find through excel.
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For 800 respondents to an inquiry of their age, results show a
mean of 42 years...
The mean age at first marriage for respondents in a survey is 23.33, with a standard...
The mean age at first marriage for respondents in a survey is 23.33, with a standard deviation of 6.13. The Z score associated with a particular age at first marriage is .35. If the proportion of the area between this particular age at first marriage and the mean is 0.14, what proportion of respondents experienced their first marriage earlier than this age? Illustrate your answer in a normal curve.
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.6...
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.69 hours, with a standard deviation of 2.49 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.07 hours, with a standard deviation of 1.54 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...
14. A random sample of 40 adults with no children under the age of 18 years results in a mean dai...
14. A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.23 hours, with a standard deviation of 2.25 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure difference in leisure time between adults with no children and adults with children (1-H2 H represent the adults with children under the age of 18 The e0% (Round...
Please show how to solve using excel. At a large university, the mean age of the...
Please show how to solve using excel.
At a large university, the mean age of the students is 22.3 years, and the standard deviation is 4 years. A random sample of 64 students is drawn. What is the probability that the average age of these students is greater than 23 years? In another large university, 25% of the students are over 21 years of age. In a sample of 400 students, what is the probability that more than 110 of...
A random sample of 40 adults with no children under the age of 18 years results...
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.55 hours, with a standard deviation of 2.28 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.14 hours, with a standard deviation of 1.86 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...
A random sample of 40 adults with no children under the age of 18 years results...
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.52 hours, with a standard deviation of 2.36 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.46 hours, with a standard deviation of 1.53 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...
The mean age of lead actresses from the top grossing movies of is 29.6 years with...
The mean age of lead actresses from the top grossing movies of is 29.6 years with a standard deviation of 6.35 years. Assume the distribution of the actresses’ ages is approximately Normal. What is the probability that 5 randomly selected actresses will have a sample mean age younger than 20 years? Show work.
Ages of Proofreaders At a large publishing company, the mean age of proofreaders is 36.2 years...
Ages of Proofreaders At a large publishing company, the mean age of proofreaders is 36.2 years and the standard deviation is 3.7 years. Assume the variable is normally distributed. a. What percentage of the company's proofreaders are between 36 and 37.5 years old? b. If a random sample of 15 proofreaders is selected, find the probability that the mean age of the proofreaders in the sample will be between 36 and 37.5 years.
A random sample of 40 adults with no children under the age of 18 years results...
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.57 hours, with a standard deviation of 241 hours. A random sample of 40 adults with children under the age of 18 results in a meandaly leisure time of 424 hours, with a standard deviation of 1.73 hours. Construct and interpreta 90% confidence interval for the mean difference in leisure time between adults with no children and...
1. In 1990, the mean height of women 20 years of age or older was 63.7...
1. In 1990, the mean height of women 20 years of age or older was 63.7 inches based on data obtained from the CDC. Suppose that a random sample of 45 women who are 20 years of age or older in 2015 results in a mean height of 63.9 inches with a standard deviation of 0.5 inch. Does your sample provide sufficient evidence that women today are taller than in 1990? Perform the appropriate test at the 0.05 level of...
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.69 hours, with a standard deviation of 2.49 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.07 hours, with a standard deviation of 1.54 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...
14. A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.23 hours, with a standard deviation of 2.25 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure difference in leisure time between adults with no children and adults with children (1-H2 H represent the adults with children under the age of 18 The e0% (Round...
Please show how to solve using excel.
At a large university, the mean age of the students is 22.3 years, and the standard deviation is 4 years. A random sample of 64 students is drawn. What is the probability that the average age of these students is greater than 23 years? In another large university, 25% of the students are over 21 years of age. In a sample of 400 students, what is the probability that more than 110 of...
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.52 hours, with a standard deviation of 2.36 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.46 hours, with a standard deviation of 1.53 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...
Ages of Proofreaders At a large publishing company, the mean age of proofreaders is 36.2 years and the standard deviation is 3.7 years. Assume the variable is normally distributed. a. What percentage of the company's proofreaders are between 36 and 37.5 years old? b. If a random sample of 15 proofreaders is selected, find the probability that the mean age of the proofreaders in the sample will be between 36 and 37.5 years.
A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.57 hours, with a standard deviation of 241 hours. A random sample of 40 adults with children under the age of 18 results in a meandaly leisure time of 424 hours, with a standard deviation of 1.73 hours. Construct and interpreta 90% confidence interval for the mean difference in leisure time between adults with no children and...
1. In 1990, the mean height of women 20 years of age or older was 63.7 inches based on data obtained from the CDC. Suppose that a random sample of 45 women who are 20 years of age or older in 2015 results in a mean height of 63.9 inches with a standard deviation of 0.5 inch. Does your sample provide sufficient evidence that women today are taller than in 1990? Perform the appropriate test at the 0.05 level of...
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