Question

example #7: masses (kilograms) andrew 2.4 2.5 3.6 3.1 2.5 2.7 maria x 3.1 2.7 2.9 3.3 2.8 andrew and maria each collected six rocks, and the masses of the rocks are shown in the table above. the mean of the masses of the rocks maria collected is 0.1 kilogram greater than the mean of the masses of the rocks andrew collected. what is the value of x? example #8: if x is the average (arithmetic mean) of m and 9, y is the average of 2m and 15, and z is the average of 3m and 18, what is the average of x, y, and z in terms of m? a) m + 6 b) m + 7 c) 2m + 14 d) 3m + 21 margin of error margin of error is usually a small amount (number or %) allowed for miscalculation or mistakes. example: “....results showed 48% of couples get divorced with a margin of error of 3%.” this means there may be some mistakes in the study to get 48%, so the “true” percentage of divorce is between 45% and 51%, outside of this confidence interval would be doubtful but not impossible. example #9 based on random samples of river heights, oceanographers estimate that 4,800 cubic kilometers (km³) of freshwater is discharged into the arctic ocean annually. the estimate has a margin of error of 240 km³ at the 90% confidence level. which of the following is the most reasonable what is the interval of acceptable volume of freshwater? example #10 in a poll of 1,578 randomly selected american adults, 44.8% of the respondents said that airlines should allow in - flight calls on airplanes. the poll reported a margin of error of 2.5% at a 95% confidence level. which of the following is most likely to be equal to the percentage of all what is the interval of acceptable percentage who believe airlines should allow inflight calls? example #11 based on carlos previous quiz scores, he estimates that he will get 86 points on his next quiz. the estimate has margin of error of 3 points at the 95% confidence interval. what is a reasonable quiz score on his next quiz? a) 82 b) 84 c) 90 d) 95 e) 98 example #12) based on the data from the study, an estimate of the percent of us teens who are heavy texters is 30% and the associated margin of error is 3%. which of the following is correct statement: a) approximately 3% of the teens in the study who are classified as heavy texters are not really heavy texters. b) it is not possible that the percent of all us teens who are heavy texters is less than 27% c) the percent of all us teens who are heavy texters is 33% d) it is doubtful that the percent of all us teens who are heavy texters is 35%

Explanation:

Example #7

Step1: Calculate Andrew's mean

Andrew's rock - masses are \(2.4,2.5,3.6,3.1,2.5,2.7\). The sum of his rock - masses is \(2.4 + 2.5+3.6 + 3.1+2.5+2.7=16.8\). The mean of Andrew's rock - masses, \(\bar{A}=\frac{16.8}{6}=2.8\) kg.

Step2: Calculate Maria's mean

Maria's rock - masses are \(x,3.1,2.7,2.9,3.3,2.8\). The sum of her rock - masses is \(x + 3.1+2.7+2.9+3.3+2.8=x + 14.8\). The mean of Maria's rock - masses, \(\bar{M}=\frac{x + 14.8}{6}\).

Step3: Set up the equation

Since the mean of the masses of the rocks Maria collected is \(0.1\) kilogram greater than the mean of the masses of the rocks Andrew collected, we have the equation \(\frac{x + 14.8}{6}=2.8 + 0.1\).

Step4: Solve the equation

First, simplify the right - hand side: \(2.8+0.1 = 2.9\). Then, multiply both sides of the equation \(\frac{x + 14.8}{6}=2.9\) by \(6\) to get \(x + 14.8=17.4\). Subtract \(14.8\) from both sides: \(x=17.4−14.8 = 2.6\).

Step1: Find the expressions for \(x\), \(y\), and \(z\)

The average \(x\) of \(m\) and \(9\) is \(x=\frac{m + 9}{2}\). The average \(y\) of \(2m\) and \(15\) is \(y=\frac{2m+15}{2}\). The average \(z\) of \(3m\) and \(18\) is \(z=\frac{3m + 18}{2}\).

Step2: Find the average of \(x\), \(y\), and \(z\)

The average of \(x\), \(y\), and \(z\) is \(\frac{x + y+z}{3}\). Substitute the expressions for \(x\), \(y\), and \(z\) into the formula:
\[

$$\begin{align*} \frac{x + y+z}{3}&=\frac{\frac{m + 9}{2}+\frac{2m+15}{2}+\frac{3m + 18}{2}}{3}\\ &=\frac{\frac{(m + 9)+(2m+15)+(3m + 18)}{2}}{3}\\ &=\frac{m + 9+2m+15+3m + 18}{6}\\ &=\frac{6m+42}{6}\\ &=m + 7 \end{align*}$$

\]

The estimate of the volume of freshwater discharged into the Arctic Ocean is \(V = 4800\) cubic kilometers, and the margin of error is \(E=240\) cubic kilometers. The interval of acceptable volume is calculated by subtracting and adding the margin of error from the estimate.
The lower - bound is \(V - E=4800−240 = 4560\) \(km^{3}\), and the upper - bound is \(V + E=4800 + 240=5040\) \(km^{3}\).

Answer:

\(2.6\)

Example #8