1 Doctors:The following table is a list ofthe number of MD’s per 100,000 people in 2002 as reported as an indicator ofthe availability of health care in the 50 states and the District of Columbia.The values are ranked from in descending order.D.C.683MAINE250MONTANA215MASSACHUSETTS427TENNESSEE250KANSAS210NEWY ORK385WASHINGTON250GEORGIA208MARY LAND378OHIO248INDIANA207CONNECTICUT360DELAWARE242TEXAS204V ERMONT346OREGON242ALABAMA202RHODEISLAND341NORTHCAROLINA241SOUTHDAKOTA201NEWJERSEY305FLORIDA237UTAH200PENNSY LV ANIA291COLORADO236ARIZONA196HAWAII280MISSOURI233ALASKA194ILLINOIS265MICHIGAN230ARKANSAS194MINNESOTA263NEBRASKA230 IOWA178LOUISIANA258NORTHDAKOTA228WY OMING176WISCONSIN256NEWMEXICO222NEV ADA174V IRGINIA253WESTV IRGINIA221MISSISSIPI171CALIFORNIA252KENTUCKY219OKLAHOMA163NEWHAMPSHIRE251SOUTHCAROLINA219IDAHO1611. Whoare the ”individuals” or ”elements” described in this data set?The number of doctors 2. Whatis the variable observed or measured to each ”individual” or ”element”?How many doctors are in each state3. Whywas this study done?To see the availability of healthcare providers 4. Whenwas the study done? In 20025. Fill in the Table of Frequencies and Percentswithin each of the intervals.IntervalFrequency% withn interval6. Drawthe Histogram of Number of Doctors per 100,000 in the space provided or copyfrom Minitab.7. Howwould you describe the shape of the distribution of the Number ofDoctors per100,000 people?8. What percent of the states have between 200 and300 doctors for every100,000 people.9. Do you think that the majority of states have More or Less than 300 doctors per 100,000 people?10. The mean x of the 51 observations is 251.9. Doyou expect the median to be Higher orLower than the mean? Why?11. The rangeis just the difference between the maximum and the minimum value. What isthe range of the data?12. Findthe Five Number Summary?13. By using the 1.5IQR Rule, List the sevenobservations(the states or D.C.) that may be classified as outliers.14. Drawthe Box-Plot of the Data either by hand or via software. Rememberto representthe outliers with either an X or an O.15. Brieflydiscuss the results above by interpreting what you believe is revealed whendiscussing the availabiliy of heath care in the 50 states and Washington D.C.2 Bar Charts: Spam EmailsHere is acompilation of the most common types of spam emails.Type of spamPercentAdult19Financial20Health7Internet7Leisure6Products25Scams91. Do the Percentages add to 100%? What Percentagewill be left for a category of Other?2. Create a Bar chart of the variable Type of Spam in the space providedeither by hand or via Minitab.3. Canwe ever discuss shape, center, or spread with a Categorical variable?Yes or No.3 Summary StatisticsHere are the lengths ofright-hand span widths (cm) of a sample of 7 students. 8.9 23 18 14 8 9.5 121. Calculatethe mean ¯x and the Median M2. Calculatethe Standard Deviation s3. Calculatethe Five Number Summary4. Whydo you believe the mean is much larger than the median?5. Whatwould happen to the values of the summary statistics if we foundout thatright-hand span of a 23 cm was a typo and should have instead been a 13 cminstead? Just say either increase, decrease, or stay the same for the eachvalue below.x¯sminQ1MQ3Max6. What would happen to the summary statistics ifeach data value wasconverted to inches? There are 2.54 cm to an inch.4 Normal Distributions4.1 Using the 68-95-99.7 RuleScores on a University exam arenormally distributed with a mean µ of68 and a standard deviation σ of 9.Use the 68-95-99.7 Rule to answer the following questions.1. Drawthe Density Curve and label ±3σ awayfrom the mean.2. Whatproportion of students score between a 59 to 77?3. Whatproportion of students score between a 50 to 86?4. Whatproportion of students score above a 86?5. What proportion of students scorebelow a 41?6. Whatproportion of students score above a 59?7. What must a student score on the exam to be inthe bottom 2.5% of scoreson the exam?4.2 Using Table A (Standard Normal Distribution) tofind Percentiles1. Find z< 1.962. Find z< −2.463. Find z> 1.504. Find z> −1.005. Find−1.25 < z < 1.754.3 Finding Proportions for a Normal Distributionwith mean µand standard deviation σ (AKA Finding zscores to use Table A)It has been shown that lengthsof pregnancies follow a Normal Distribution with mean µ of 266 days and standard deviation σ of 16 days. Use this information to answer the followingquestions.1. Drawthe Density Curve for Lengths of Pregnancies.2. Whatproportion of pregnancies last less than 275 days?3. Whatproportion of pregnancies last more than 275 days?4. Whatproportion of pregnancies last between 245 days and 280 days?5. Whatproportion of pregnancies last less than 290 days?6. What proportion of pregnancies last more than255 days?
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