Math 221 – Elementary Statistics



MATH211 Statistics—Lab Week 4Questions 1 through 7 worth 6 points; question 8 worth 8 points: total of 50 points for labName: _______________________Statistical Concepts:68-95-99.7% RuleComparing Normal DistributionsCentral Limit TheoremBasic ProbabilityContingency Tables68-95-99.7% RuleUsing the class data set:From the Week 2 Lab: The next two questions need the Height mean and standard deviation for the males and the females: variable Height by Gender.[Repeat of directions from Week 2 Lab: Calculate descriptive statistics for the variable Height by Gender. Click on Insert and then Pivot Table. Click in the top box and select all the data (including labels) from Height through Gender. Also click on “new worksheet” and then OK. On the right of the new sheet, click on Height and Gender, making sure that Gender is in the Rows box and Height is in the Values box. Click on the down arrow next to Height in the Values box and select Value Field Settings. In the pop-up box, click Average. Type in the averages below. Then click on the down arrow next to Height in the Values box again and select Value Field Settings. In the pop-up box, click on StdDev. Type the standard deviations below.] Include units in your answers. (No points for this answer as from previous lab.)MeanStandard deviationFemales67.06 inches3.11 inchesMales69.67 inches3.31 inchesUsing the empirical rule, 95% of female heights should be between what two values? Either show your work or explain how you calculated your answer. Include units in your answers.67.06 – 2(3.11) = 60.84 67.06 + 2(3.11) = 73.28 95% of female heights are expected to fall between 60.84 inches and 73.28 inches.Using the empirical rule, 68% of male heights should be between what two values? Either show your work or explain how you calculated your answer. Include units in your answers.69.67 – 3.31 = 66.3669.67 + 3.31 = 72.9868% of male heights are expected to fall between 66.36 inches and 72.98 paring Normal DistributionsSay you select one female with a height that is 65.5 inches. Using the formula for z scores, what is the z score for this person? Second, say that you select one male with a height of 71.9 inches. Using the formula for z scores, what was the z score for this person?Calculate the two z-scores. Female z-score = (65.5 – 67.06) / 3.11 = -0.502Male z-score = (71.9 – 69.67) / 3.31 = 0.674Using the z-scores or the respective probability, explain which person (selected female or selected male) is taller than average. The male is taller than average since that person’s z score is positive. Any positive z score is above the average while any negative z score is below the average.Basic ProbabilityCreate a Decision Tree with all outcomes for the event of flipping 2 fair coins and rolling a 9-sided die.Coin1 Coin2 Die12345678H9HT12345678912345678H9TT123456789Set of all outcomes ={HH1,HH2,HH3,HH4,HH5,HH6,HH7,HH8,HH9, HT1,HT2,HT3,HT4,HT5,HT6,HT7,HT8,HT9, TH1,TH2,TH3,TH4,TH5,TH6,TH7,TH8,TH9, TT1,TT2,TT3,TT4,TT5,TT6,TT7,TT8,TT9}36 total outcomesWhat is the probability of flipping two tails and rolling a 6?P(TT6) = 1/36 or 2.78%What is the probability of first flipping a head, and then flipping a tail, and finally rolling a 2? P(HT2) = 1/36 or 2.78%Contingency Table ProbabilityUsing the data provided, count how many respondents fall into each box of the following contingency table. Then, use this table to answer the following probability questions. Excel does not have a way to create the table automatically. You can sort the data based on TV hours and then count how many respondents fall into each box below. Remember that the total of the boxes (number in the lower right corner) should equal the total number of respondents to the survey. 1 to 2 hours of TV3 to 4 hours of TV5 or more hours of TVTotalsCoin of 2 or 346010Coin of 4 or 557517Coin of 6 or 71618Totals1019635Give the respective Probabilities:P(1 to 2 hours of TV) 10/35 = 28.6%P(Coin of 4 or 5) 17/35 = 48.6%P(1 to 2 hours of TV | Coin of 2 or 3) 4/10 = 40%P(Coin of 6 or 7 | 3 to 4 hours of TV) 6/19 = 31.6%P(Coin of 6 or 7 AND 5 or more hours of TV) 1/35 = 2.9%P(Coin of 6 or 7 OR 5 or more hours of TV) 13/35 = 37.1%P(1 to 2 hours of TV AND 3 to 4 hours of TV) 0%P(1 to 2 hours of TV OR 3 to 4 hours of TV) 29/35 = 82.9% ................
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