A survey found that women's heights are normally distributed with mean 63.4 in. and standard deviation 2.4 in. The survey also found that men's heights are normally distributed with a mean 68.1 in. and standard deviation 2.8 . Complete parts a through c below. a. Most of the live characters at an amusement park have height requirements with a minimum of 4 ft 9 in . and a maximum of 6 ft 2 in . Find the percentage of women meeting the height requirement. The percentage of women who meet the height requirement is \( 99.62 \% \). (Round to two decimal places as needed ) b. Find the percentage of men meeting the height requirement. The percentage of men who meet the height requirement is \( 98.26 \% \). (Round to two decimal places as needed) c. If the height requirements are changed to exclude only the tallest \( 5 \% \) of men and the shortest \( 5 \% \) of women, what are the new height requirements? The new height requirements are at least \( \square \) in and at most \( \square \) in]. (Round to one decimal place as needed)

We measured the flat area of 30 apartments and we have measured the following values in \( \mathrm{m}^{2} \) : \( 82.6,57.3,70.4,65,48.4,103.8,73.6,43.5,66.1,93,52.6,70,84.2,55,81.3,61.5,75.1,34.8 \), \( 62.4,116,70.1,63.6,93,59.2,65.9,77.2,52.8,68.7,79.2,87.4 \). a) Create a table of grouped frequency distribution for the number of classes \( k=9 \). b) Construct a histogram of relative frequencies of the flat areas. c) From the specified values estimate the sample mean and the variance. d) From the middle values of the intervals and from their frequencies estimate the sample mean and the variance.

6. A linear regression model relates the income of a dentist to the number of patients seen. What can be extrapolated from this information? the income of all doctors the income of all optometrists the income of all patients that regularly see a dentist the income of all dentists

magazine provided results from a poll of 1000 adults who were asked to identify their favorite pie. Among the 1000 respondents, \( 11 \% \) chose chocolate pie, and the margin of error was given as \( \pm 5 \) percentage points. Describe \( w \) eant by the statement that "the margin of error was given as \( \pm 5 \) percentage points." A. The statement indicates that the interval \( 11 \% \pm 5 \% \) is likely to contain the true population percentage of people that prefer chocolate pie. B. The statement indicates that the study is only \( 5 \% \) confident that the true population percentage of people that prefer chocolate pie is exactly \( 11 \% \). C. The statement indicates that the true population percentage of people that prefer chocolate pie is in the interval \( 11 \% \pm 5 \% \). D. The statement indicates that the study is \( 100 \%-5 \%=95 \% \) confident that the true population percentage of people that prefer chocolate pie is \( 11 \% \).

A magazine provided results from a poll of 500 adults who were asked to identify their favorite pie. Among the 500 respondents, \( 13 \% \) chose chocolate pie, and the margin of error was given as \( \pm 4 \) percentage points. What values do \( \hat{p}, \hat{q}, \mathrm{n} \), E , and p represent? it the confidence level is \( 99 \% \), what is the value of \( \alpha \) ? The value of \( \hat{p} \) is The value of \( \hat{q} \) is The value of n is The value of E is The value of p is If the confidence level is \( 99 \% \), what is the value of \( \alpha \) ? \( \alpha=\square \) (Type an integer or a decimal. Do not round.)

Here are summary statistics for randomly selected weights of newborn girls: \( \mathrm{n}=188, \overline{\mathrm{x}}=29.3 \mathrm{hg}, \mathrm{s}=6.7 \mathrm{hg} \). Construct a confidence interval estimate of the mean. Use a \( 95 \% \) confidence level. Are these results very different from th confidence interval \( 27.5 \mathrm{hg}<\mu<30.1 \mathrm{hg} \) with only 15 sample values, \( \bar{x}=28.8 \mathrm{hg} \), and \( s=2.4 \mathrm{hg} \) ? What is the confidence interval for the population mean \( \mu \) ? \( \square \mathrm{hg}<\mu<\square \mathrm{hg} \) (Round to one decimal place as needed.) Are the results between the two confidence intervals very different? A. No, because each confidence interval contains the mean of the other confidence interval. B. Yes, because the confidence interval limits are not similar. C. No, because the confidence interval limits are similar. D. Yes, because one confidence interval does not contain the mean of the other confidence inferval.