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Elementary Statistics Supplementary Notes

David W. Lyons

Section 3.5 Sampling Problems

Subsection 3.5.1 Sum of Draws Practice Problems (Ch 16–18)

Exercises Exercises

Problems 1–10 below are about the following game. A fair die is rolled 600 times. On each roll, you win $4 if you get a 6. You lose $1 if you get something different from a 6.
1.
The total winnings will be about $ give or take $ or so.
Answer.
The box model has six tickets: (1) 4 and (5) \(-1\)’s.
\begin{align*} \AVE(\text{box}) \amp = -1/6\\ \SD(\text{box}) \amp = 5\sqrt{\frac{1}{6}\cdot \frac{5}{6}} \approx 1.86\\ \text{expected(sum) }\amp = -100\\ \text{SE(sum) } \amp \approx 45.64 \end{align*}
2.
The chance of breaking even or better is about .
Answer.
\begin{align*} z \amp \approx \frac{0 - (-100)}{45.64} \approx 2.19\\ P\amp \approx 1.5\% \end{align*}
3.
The number of sixes rolled will be about give or take .
Answer.
The box model has (1) 1 and (5) 0’s.
\begin{align*} \AVE(\text{box}) \amp = 1/6\\ \SD(\text{box}) \amp = \sqrt{\frac{1}{6}\cdot \frac{5}{6}} \approx .373\\ \text{expected(sum) } \amp = 100\\ \text{SE(sum) }\amp \approx 9.13 \end{align*}
4.
The chance that the number of sixes will be ten or more above the expected number is about .
Answer.
\begin{align*} z \amp \approx \frac{110 - 100}{9.13} \approx 1.1\\ P \amp \approx 14\% \end{align*}
5.
The sum of the numbers on the die face on all the rolls will be about give or take .
Answer.
The box model has the tickets 1,2,3,4,5,6.
\begin{align*} \AVE(\text{box}) \amp = 3.5\\ \SD(\text{box}) \amp \approx 1.71 \text{ (no shortcut!!)}\\ \text{expected(sum) } \amp = 2100\\ \text{SE(sum) } \amp \approx 41.83 \end{align*}
6.
The chance that the sum will be in the range 2050--2150 is about .
Answer.
\begin{align*} z \amp \approx \frac{2150 - 2100}{41.83} \approx 1.2\\ P \amp \approx 77\% \end{align*}
7.
There is about a 50% chance that the sum will be in the range 2100 plus or minus .
Answer.
\begin{gather*} z \text{ for 50% is about } .67\\ \text{range is } 2100 \pm 28 \end{gather*}
8.
The total number of even rolls will be about give or take .
Answer.
One possible box model has (3) 1’s and (3) 0’s. A simpler box model has (1) 1 and (1) 0 with equal probabilities of 1/2 each.
\begin{align*} \AVE(\text{box}) \amp = 1/2\\ \SD(\text{box})\amp = 1/2\\ \text{expected(sum) } \amp = 300\\ \text{SE(sum) } \amp \approx 12.25 \end{align*}
9.
The chance that the number of even rolls will be off by 20 or more from the expected number is about .
Answer.
\begin{align*} z \amp \approx \frac{320 - 300}{12.25} \approx 1.63\\ P \amp \approx 10\% \end{align*}
10.
On every 6th roll (that is, on roll numbers 6, 12, 18, etc) you win a blue marble if exactly half of the last 6 rolls came up even. The chance that you win 35 or more blue marbles in 600 rolls is . Use the continuity correction to be as precise as possible.
Answer.
The probability of exactly 3 evens in 6 rolls is \({6 \choose 3}\left(\frac{1}{2}\right)^6 = 31.25\%\) (using the binomial probability formula). A box model has 3125 1’s and 6875 0’s, with 100 draws. The average of the box is \(.3125\) and the SD of the box is \(\sqrt{.3125\cdot .6875} \approx .464\text{.}\) The expected sum is \(31.25\) and the SE for the sum is \(4.64\text{.}\) The \(z\) value for 35 blues is
\begin{equation*} z = \frac{34.5-31.25}{4.64}\approx .70. \end{equation*}
The normal curve table has an area of about \(52\%\) for \(z=.70\text{,}\) so the probability is about \((100-52)/2=24\%\) (if you do not use the continuity correction, you get \(z=.81\) and a probability of about \(21\%\)).
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