This algebra 2 normal worksheet answer key solves all 22 problems of the data, normal distributions and statistical inference worksheet step by step.
Every solution shows the formula, the substitution and the final result in bold. Table values are rounded to four decimals, so a calculator may differ in the last digit. Study descriptions can be worded differently as long as they include random selection or random assignment where required.
Algebra 2 Normal Worksheet Answer Key: problems 1–5
Problem 1: Find a mean
Add the scores: \(12+15+9+20+14=70\).
Divide by 5: \(\overline{x}=\frac{70}{5}=14\).
Answer: 14 points per game.
Problem 2: Standard deviation by hand
Mean: \(\frac{2+4+4+4+5+5+7+9}{8}=\frac{40}{8}=5\).
Squared deviations: \(9,1,1,1,0,0,4,16\), with sum \(32\).
\(\sigma=\sqrt{\frac{32}{8}}=\sqrt{4}=2\).
Answer: mean 5 minutes, \(\sigma=\)2 minutes.
Problem 3: Standard deviation with a root
The mean is \(\frac{50}{5}=10\).
Squared deviations: \(16,4,0,4,16\), sum \(40\).
\(\sigma=\sqrt{\frac{40}{5}}=\sqrt{8}\approx 2.83\).
Answer: \(\sigma\approx 2.83\).
Problem 4: Shift and scale
Adding a constant shifts the mean and does not change the spread; multiplying by \(k\) multiplies both.
- Mean \(40+5=45\); standard deviation stays \(6\). Answer: \(\$45\) and \(\$6\).
- Mean \(2\times 40=80\); standard deviation \(2\times 6=12\). Answer: \(\$80\) and \(\$12\).
Problem 5: Compare two spreads
Town A: the sum is \(300\), so the mean is \(50\). Squared deviations \(9,1,0,0,1,9\) sum to \(20\), so \(\sigma=\sqrt{\frac{20}{6}}\approx 1.83\).
Town B: the sum is \(300\), so the mean is \(50\). Squared deviations \(100,36,0,0,36,100\) sum to \(272\), so \(\sigma=\sqrt{\frac{272}{6}}\approx 6.73\).
Answer: both means are \(50^{\circ}\text{F}\), but Town B is more variable (\(6.73 > 1.83\)).
Algebra 2 Normal Worksheet Answer Key: problems 6–10
Problem 6: Intervals from the 68-95-99.7 rule
Go \(1\), \(2\) and \(3\) standard deviations (\(15\), \(30\), \(45\)) from \(100\).
\(68\%\): \(100\pm 15\) gives \(85\) to \(115\).
\(95\%\): \(100\pm 30\) gives \(70\) to \(130\).
\(99.7\%\): \(100\pm 45\) gives \(55\) to \(145\).
Answer: \(85\)–\(115\), \(70\)–\(130\), \(55\)–\(145\).
Problem 7: Apply the rule to heights
\(59=64-2(2.5)\) and \(69=64+2(2.5)\), so the interval is \(\pm 2\sigma\).
- About \(95\%\).
- \(66.5=64+2.5\) is one standard deviation above the mean. Half the data is above the mean, and \(34\%\) is between \(\mu\) and \(\mu+\sigma\), so \(50\%-34\%=16\%\). Answer: about \(16\%\).
- \(56.5=64-3(2.5)\), so it is \(3\sigma\) below the mean. Outside \(\pm 3\sigma\) lies \(0.3\%\), split into two tails, so \(0.15\%\) is below. Answer: about \(0.15\%\).
Problem 8: Waiting times
\(26=30-4\), \(38=30+2(4)\) and \(42=30+3(4)\).
- From \(-1\sigma\) to \(+2\sigma\): \(34\%+34\%+13.5\%=81.5\%\). Answer: about \(81.5\%\).
- Above \(+3\sigma\) lies \(0.15\%\). Answer: about \(0.15\%\).
Problem 9: Compute a z-score
\(z=\frac{85-70}{6}=\frac{15}{6}=2.5\).
Answer: \(z=2.5\); Ethan’s score is \(2.5\) standard deviations above the mean.
Problem 10: Negative z-score
\(z=\frac{41-50}{4.5}=\frac{-9}{4.5}=-2\).
Answer: \(z=-2\); this delivery was \(2\) standard deviations faster than average.
Algebra 2 Normal Worksheet Answer Key: problems 11–15
Problem 11: Value from a z-score
Use \(x=\mu+z\sigma\).
- \(x=68+(-0.8)(5)=68-4=64\). Answer: 64.
- \(x=68+(1.4)(5)=68+7=75\). Answer: 75.
Problem 12: Compare two scores
Mei: \(z=\frac{78-70}{5}=1.6\). Noah: \(z=\frac{84-75}{7}\approx 1.29\).
Answer: Mei did better relative to her class, since \(1.6 > 1.29\).
Problem 13: Area to the left
- The table value for \(1.25\) is \(0.8944\), about \(89.4\%\).
- The table value for \(-0.75\) is \(0.2266\), about \(22.7\%\); it is less than \(0.5\) because \(z\) is negative.
Problem 14: Area between two z-scores
Subtract the left areas: \(0.9332-0.3085=0.6247\).
Answer: 0.6247, about \(62.5\%\) of the data.
Problem 15: Name the study
- Residents are randomly selected and asked a question: sample survey.
- Treatments are randomly assigned by the researcher: experiment.
- Groups formed on their own and nothing is assigned: observational study.
Only the experiment can support a conclusion about cause and effect.
Algebra 2 Normal Worksheet Answer Key: problems 16–20
Problem 16: Spot the bias
Students at a basketball game are more likely to enjoy sports, so they overrepresent students who favor more PE. Not every student had an equal chance of being asked.
Better method: number the students \(1\) to \(800\), use a random number generator to pick \(50\) different numbers, and survey exactly those students.
Problem 17: Design an experiment
Number the classmates \(1\) to \(30\) and use a random number generator to choose \(15\) for the music group; the other \(15\) form the no-music (control) group. Both groups study the same material for the same time, then take the same quiz.
Compare the two mean quiz scores. If the difference is larger than random assignment alone would usually produce, the music is a plausible cause. Because the groups were assigned randomly, a cause-and-effect claim is justified for these \(30\) students; generalizing further needs a random sample of students.
Problem 18: Interpret a poll
- \(52\%\pm 3\%\) gives \(49\%\) to \(55\%\).
- No. The interval contains values below \(50\%\) (for example \(49\%\)), so a majority is possible but not established.
Problem 19: Exam percentiles
- \(90=72+2(9)\), so \(z=2\). The area to the left is \(50\%+34\%+13.5\%=97.5\%\). Answer: about the 97.5th percentile.
- \(63=72-9\), so \(z=-1\). The area to the left is \(50\%-34\%=16\%\). Answer: about the 16th percentile.
- \(100\%-97.5\%=2.5\%\) and \(0.025\times 400=10\). Answer: about 10 students.
Problem 20: Filling machine
- \(15.6=16.0-2(0.2)\), so \(z=-2\). About \(2.5\%\) of bags.
- \(16.6=16.0+3(0.2)\), so \(z=3\). \(0.15\%\times 2000=3\). About 3 bags.
- \(16.0\pm 2(0.2)\) gives 15.6 oz to 16.4 oz.
Algebra 2 Normal Worksheet Answer Key: problems 21–22
Problem 21: Percentile of a test score
- \(z=\frac{1310-1050}{200}=1.3\). Answer: \(z=1.3\).
- The table value for \(1.3\) is \(0.9032\), so he scored above about \(90\%\) of test takers.
- \(x=1050+(-0.5)(200)=950\). Answer: 950.
Problem 22: Margin of error by simulation
- Counting from the left, the bars below \(26\%\) contain 5 samples, and the bars above \(46\%\) contain 5 samples. The middle \(95\%\) runs from \(26\%\) to \(46\%\).
- \(\frac{46-26}{2}=10\). Answer: about 10 percentage points.
- \(41\%\pm 10\%\) gives \(31\%\) to \(51\%\). Since \(31 < 35 < 51\), \(35\%\) is plausible.
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