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WEEK 4 HOMEWORK:  LANE CHAPTER 7 AND ILLOWSKY CHAPTERS 6 AND 7

THE NORMAL DISTRIBUTION Z-TABLES ARE INCLUDED IN THE COURSE RESOURCES AND YOU ARE TO USE THEM RATHER THAN SOFTWARE TO SOLVE THESE PROBLEMS.  THIS IS STRAIGHT FORWARD TABLE READING. 

THIS WEEK’S CONCEPTS ARE REALLY THE HEART OF OUR COURSE.  PROBABILITY (FROM LAST WEEK) IS THE DRIVING FORCE BEHIND STATISTICS (AND THEORETICAL PHYSICS – WATCH THE PBS SHOW “PARTICLE FEVER”). 

THE NORMAL DISTRIBUTION AND THE AREAS UNDER PARTS OF IT ARE ALL WE ARE TRYING TO FIGURE OUT IN STATISTICS.  THESE WOULD BE THE AREAS IN THE TABLE THAT CORRESPOND TO SPECIFIC (CALCULATED) Z-VALUES.  OUR DATA ARE THE X-VALUES AND WE CALCULATE OUR Z-VALUES FROM THEM.  WE CAN ALSO BACK-CALCULATE X-VALUES FROM Z-VALUES. 

LONG STORY SHORT:  THERE ARE SPECIFIC Z-VALUES OF INTEREST REFERRED TO AS “CRITICAL VALUES” AND THOSE ARE THE ONES THAT CORRESPOND TO SMALL (RARE) AREAS IN ONE OR BOTH “TAILS” OF OUR NORMAL DISTRIBUTION:  1%, 5% OR 10% OF THE TOTAL AREA UNDER THE NORMAL CURVE.   THESE ALREADY SMALL AREAS CAN ALSO BE SPLIT BETWEEN BOTH ENDS LEAVING 0.5%, 2.5% AND 5% IN EACH TAIL.  THE PERCENTAGES ARE OUR “SIGNIFICANCE LEVELS” AND WE CHOSE ONE BASED ON HOW SURE WE WANT TO BE OF OUR CONCLUSION:  90%, 95% OR 99% (CORRESPONDING TO THE AREAS OF 10%, 5% AND 1%)

NON-TEXT PROBLEM #1:  LOOK THESE Z-VALUES UP FOR THESE AREAS:  + 1%, + 5%, + 10% AND + 0.5%, + 2.5%% (AND YOU ALREADY HAVE THE + 5%) AND WRITE THEM DOWN AS THEY DON’T CHANGE.  THE Z-VALUES ON THE LEFT END ARE NEGATIVE AND THOSE ON THE RIGHT END ARE POSITIVE. 

HERE IS HOW WE USE THESE Z-VALUES TO SEE IF OUR DATA ARE IN THE “RARE” OR “UNUSUAL” AREAS TO THE FAR LEFT (OR RIGHT) OF OUR NORMAL DISTRIBUTION.  WHY DO WE CARE IF DATA ARE RARE?  YOU WILL SEE. 

LET’S START WITH OUR QUANTITATIVE, CONTINUOUS DATA VALUES (REFERRED TO AS THE X-VALUES).  WE NEXT CALCULATE A Z-VALUE FROM OUR X-VALUE:  [Z = (X – MEAN) / STD DEV ].  THIS Z-VALUE IS SIMPLY THE NUMBER OF STANDARD DEVIATIONS OUR X-VALUE IS FROM THE MEAN. 

NOW, WE COMPARE OUR CALCULATED Z-VALUE (REFERRED TO AS THE “TEST STATISTIC”) TO OUR CRITICAL VALUE (FOR WHATEVER SIGNIFICANCE LEVEL WE CHOSE). 

IF THE TEST STATISTIC IS GREATER THAN THE POSITIVE (+) CRITICAL Z-VALUE WE ARE IN THE “UNUSUAL” OR RARE AREA IN THE RIGHT TAIL OF THE NORMAL DISTRIBUTION.  IF IT IS LESS THAN THE NEGATIVE (–) CRITICAL VALUE IT ALSO IN THE RARE AREA IN THE LEFT TAIL. 

THIS IS THE WAY WE TEST A HYPOTHESIS, AS YOU WILL LEARN IN LATER CHAPTERS.  IF OUR TEST STATISTIC ENDS UP IN THE RARE AREA, WE REJECT OUR HYPOTHESIS, IF NOT WE ACCEPT IT.  BUT, KEEP IN MIND THAT “ACCEPTING’ DOES NOT MEAN “PROVING”.  STATISTICS PROVES NOTHING BY ITSELF.  IT SIMPLY ADDS SUPPORT TO OTHER INFORMATION. 

LET’S GET BACK TO THE NORMAL DISTRIBUTION.  THE EXCEL PAGE ATTACHED PROVIDES AN EXAMPLE OF SUCH A DISTRIBUTION.  MAKE SURE IT MAKES SENSE TO YOU.   

NOW, LET’S MOVE ON TO THIS WEEK’S HOMEWORK:  LANE (C7) FIRST

PROBLEM #8  THIS IS A STRAIGHT FORWARD Z CALCULATION.  THIS Z-VALUE (SCORE) TELLS YOU HOW MANY STANDARD DEVIATIONS A SPEED OF 65 IS FROM THE MEAN (IN THIS CASE IT’S A NEGATIVE DISTANCE, MEANING IT’S BELOW THE MEAN).  USE THE CORRECT TABLE (NOT SOFTWARE) TO FIND THIS AREA (NEGATIVE - Z-VALUES) AND REMEMBER TOO THAT THE TABLES GIVE THE AREAS TO THE LEFT OF THAT Z VALUE.  TO GET THE AREAS TO THE RIGHT (BUT NOT NEEDED IN THIS PROBLEM) YOU SIMPLY SUBTRACT THE AREA TO THE LEFT FROM 1.00, SINCE THE AREA UNDER THE NORMAL CURVE ACCOUNTS FOR 100% OR 1.00 OF THE DATA. 

KEEP IN MIND THAT THESE “AREAS” ARE THE PERCENTAGE OF DATA BELOW YOUR SPECIFIC X-VALUE AND ALSO THE PROBABILITY (RELATIVE FREQUENCY) OF DATA BEING IN THAT AREA.  THIS IS THE HEART OF IT.

 AS AN EXAMPLE, TO READ THE TABLE FOR A Z-SCORE OF +1.31 GO TO THE TABLE OF POSITIVE Z-SCORES AND FIND 1.3 IN THE LEFT COLUMN AND THEN GO OVER TO THE 0.01 ON THE TOP AND GET THE READING FROM THE TABLE.  IT’S 0.9049.  IF YOU HAVE A NEGATIVE Z-SCORE USE THE OTHER TABLE.  WHAT IF YOU HAVE 1.315?  GET THE AREAS FOR 1.31 AND 1.32 AND IN OUR CASE SIMPLY APPROXIMATE THE VALUE BETWEEN THEM. 

BACK TO THE AREA OF 0.9049, THIS AREA MEANS THAT 90.49% OF OUR DATA ARE BELOW THE DATA POINT THAT HAS A CALCULATED Z-VALUE OF 1.31.  IT ALSO MEANS THAT THERE IS A 90.49% PROBABILITY THAT A DATA POINT WILL BE IN THAT AREA BELOW OUR DATA VALUE.  CONVERSELY, ABOUT 9.5% WILL BE ABOVE IT (100% - 90.49% = 9.5%).  MAKE SENSE?

PART B IN THIS PROBLEM WE USE AN EVEN SMALLER PERCENTAGE (LESS THAN 1%, BUT YOU NEED TO CALCULATE AND FIND THE AREA UNDER THE CURVE)

PART C IS A LITTLE TRICKIER.  NOTE THAT THE PROBLEM STATES THAT ONLY 10% BE OVER THE SPEED LIMIT.  OUR TABLES GIVE PERCENTAGE UNDER A SPECIFIC Z-VALUE.  SO, IF 10% ARE OVER, THAN MEANS THAT 90% ARE UNDER.  FIND THE Z-VALUE IN THE TABLE (+TABLE) CORRESPONDING (NEAREST NUMBER) TO 90%.  ONCE YOU HAVE THAT Z-VALUE YOU CAN BACK-CALCULATE TO THE ACTUAL X-VALUE, WHICH IS THE SPEED OF THE CAR.  (THE FORMULA IS THE Z = (X-MEAN)/STD DEV SO USE ALGEBRA TO SOLVE FOR “X”) 

PART D – THE DISTRIBUTION (SHAPE OF THE CURVE) IS ACTUALLY SKEWED NOT BELL SHAPED, TELL ME IF IT’S A POSITIVE OR NEGATIVE SKEW.

PROBLEM #11:  ANOTHER TRICKY ONE.  THE “TOP 30%” ARE THE ONES TO THE RIGHT SO WE NEED THE Z-VALUE THAT CORRESPONDS TO THE 70% OF THE AREA BEING BELOW IT.  ONCE YOU HAVE FOUND THAT Z-SCORE, BACK-CALCULATE THE ACTUAL X-VALUE, WHICH IS THE NECESSARY SCORE ASKED FOR.  USE THE SAME PROCEDURE FOR PART B (5% ABOVE MEANS 95% BELOW)

PROBLEM #12:  PARTS “a” IS FAIRLY STRAIGHT FORWARD, BUT PART “b” IS VERY TEDIOUS.  THIS PROBLEM INVOLVES USING THE NORMAL DISTRIBUTION (THE BELL CURVE ) WHICH IS CONTINUOUS TO APPROXIMATE THE BINOMIAL WHICH IS DISCRETE.  REMEMBER THAT A BINOMIAL HAS JUST TWO OPTIONS (LIKE HEADS/TAILS).  THE NORMAL DEALS WITH CONTINUOUS DATA. 

WE WANT THE PROBABILITY OF GETTING 15 TO 18 HEADS OUT OF 25 COIN FLIPS.  IF YOU LOOK AT THE HISTOGRAM THAT HAS THE NORMAL BELL SHAPE YOU SEE THAT EACH COLUMN IS CENTERED ON THE 1, 2, 3 ETC.  THIS MEANS THAT FOR A VALUE OF 2 THE COLUMN COVERS THE RANGE OF 1.5 TO 2.5.  DO YOU SEE THIS?  OK SO HERE IS WHAT WE DO TO USE THE NORMAL TO APPROXIMATE THE BINOMIAL 

FIRST:  WE NEED TO CALCULATE THE MEAN, WHICH HERE EQUALS THE TOTAL NUMBER OF DATA POINTS (IN THIS CASE 25 TOSSES) TIMES THE PROBABILITY OF GETTING A PARTICULAR RESULT (LIKE A HEAD), IN THIS CASE 50/50 OR 0.5.  THE VARIANCE EQUALS THE TOTAL NUMBER OF DATA POINTS TIMES THE FIRST PROBABILITY (GETTING HEADS), WHICH IS 0.5 TIMES THE PROBABILITY OF NOT GETTING HEADS (TAILS IS ALSO 0.5)  YOU CAN SEE THAT WE COULD HAVE DIFFERENT PROBABILITIES, FOR EXAMPLE IF THE COIN WERE WEIGHTED SO IT FAVORED HEADS, BUT THAT IS NOT THE CASE HERE.  ONCE WE CALCULATE THIS VARIANCE, THE STANDARD DEVIATION IS SIMPLY THE SQUARE ROOT OF THE VARIANCE.  WE NOW HAVE OUR SAMPLE MEAN AND STANDARD DEVIATION FOR OUR BINOMIAL.

TO GET PROBABILITIES FROM OUR TABLES WE NEED Z-VALUES.  WE WANT THE PROBABILITY OF GETTING 15 TO 18 HEADS OUT OF 25 TOSSES.  BUT, SINCE THE NORMAL DISTRIBUTION IS CONTINUOUS, THE 15 GOES FROM 14.5 TO 15.5 AND 18 GOES FROM 17.5 TO 18.5.  SO, OUR RANGE WHEN USING THE NORMAL TO APPROXIMATE THIS BINOMIAL IS 14.5 UP TO ____?   NOW, CALCULATE THE Z-VALUES FOR THESE TWO LIMITS. 

FROM THESE Z-VALUE WE GO TO THE TABLE AND FIND THE AREAS TO THE LEFT CORRESPONDING TO THEM.  SINCE WE ONLY WANT THE AREA BETWEEN THESE TWO LIMITS WE MUST SUBTRACT THE LOWER AREA FROM THE UPPER AREA.  IS THIS CLEAR TO YOU?  LOOK AT TEXT PICTURES TO SEE WHAT IS GOING ON. 

PART B:  INVOLVES THE ACTUAL BINOMIAL EQUATION.  THE BINOMIAL GRAPH IS NOT CONTINUOUS BUT WILL HAVE A COLUMN FOR EACH  POSSIBILITY, IN THIS CASE 15, 16, 17, AND 18  HEADS.  WE NEED TO USE THE COMPLEX BINOMIAL FORMULA THAT WE WORKED OUT IN WEEK 3 FOR PROBLEM LANE CHAPTER 5 #25.  CHECK IT OUT. 

WE MUST USE THAT FORMULA TO GET THE PROBABILITY FOR EACH OF THESE FOUR POSSIBILITIES AND THEN ADD THEM UP.  IF YOU WANT TO DO THE MATH (CORRECTLY) ON THIS ONE FOR 0.5 BONUS POINTS, SHOW ME YOUR SET UPS AND CALCULATIONS AND TOTAL  (IT’S 0.2049).  

ANSWER THIS FOR PART B:  HOW DOES THE TRUE BINOMIAL CALCULATION RESULT OF 0.2049 COMPARE TO THE NORMAL APPROXIMATION OF THE BINOMIAL (FROM PART A)?  BY WHAT PERCENTAGE IS THE APPROXIMATION OFF? 

 

ILLOWSKY CHAPTER 6

PROBLEM #60   THE EASY SOLUTION IS THAT SINCE THIS IS A NORMAL DISTRIBUTION, THE MEAN, MEDIAN AND MODE ARE THE SAME NUMBER.  SO, IF THE MEAN IS 5.3, THEN THE MEDIAN IS 5.3.  BUT, I WANT YOU TO DO THIS USING THE Z-TABLES.  START WITH THE DEFINITION OF THE “MEDIAN”.  WHAT PERCENT OF OUR DATA POINTS ARE ABOVE AND BELOW THE MEDIAN?  USING THE PERCENTAGE BELOW (WHICH IS THE AREA TO THE LEFT) GO TO THE TABLE AND GET THE Z-VALUE THAT CORRESPONDS TO THAT AREA.  USING THAT Z-VALUE AND THE MEAN AND SD, BACK CALCULATE THE X-VALUE.

PROBLEM #66:  USE THE FORMULA FOR Z:   Z = (X-u)/s  AND EXPLAIN WHAT THE RESULT MEANS.

PROBLEM #76:  DO THE MATH AND READ THE TABLE NOT SOFTWARE.  SKIP “a”

(b)  IT ASKS FOR “LESS” THAN SO IT’S THE VALUE (AREA)  IN THE TABLE.  CALCULATE THE Z-SCORE AND READ THE AREA FROM THE TABLE (WHICH IS THE PROBABILITY OR PERCENTAGE BELOW 220).

(c)  THE 80TH PERCENTILE SIMPLY MEANS 80% OF THE AREA (TO THE LEFT).  THIS MEANS THAT 80% OF THE BASEBALLS TRAVEL LESS THAN THIS DISTANCE.  FIND THE Z-VALUE CORRESPONDING TO 0.80 IN THE TABLE AND BACK-CALCULATE THE X-VALUE (THE DISTANCE).  NO SOFTWARE.

PROBLEM #88:  TRICKY.  CALCULATE THE Z-SCORE FOR 30%.  FIND THE AREA IN THE TABLE THAT CORRESPONDS TO THAT Z-SCORE.  ”AT LEAST 30” MEANS THAT WE DON’T WANT THE AREA TO THE LEFT WHICH WOULD BE THOSE SCORES LESS THAN 30%, SO WE WANT THE AREA TO THE RIGHT (THOSE SCORES GREATER THAN 30%).  SO, SINCE OUR TABLES ONLY GIVE AREAS TO THE LEFT, WE SUBTRACT IT FROM 1.00 (OR 100%).  PIECE OF CAKE. 

(b)  95TH  PERCENTILE MEANS THAT 95% OF OUR DATA ARE IN THIS AREA.  FIND THE Z-VALUE THAT CORRESPONDS TO THAT AREA.  CONVERT IT TO A X-VALUE WHICH IN THIS PROBLEM IS ALSO A PERCENTAGE SO IT’S A LITTLE CONFUSING. 

ILLOWSKY CHAPTER 7

PAGE 375 DISCUSSES THE CENTRAL LIMIT THEOREM AND SHOWS THE FORMULAS WE WILL BE USING (NOT SOFTWARE). IN ESSENCE THE CLT SAYS THAT IF WE TAKE THE MEANS OF MULTIPLE SAMPLES FROM THE SAME POPULATION, THOSE MEANS WILL BE NORMALLY DISTRIBUTED REGARDLESS OF THE ACTUAL SHAPE OF THE ACTUAL POPULATION’S DISTRIBUTION.. 

PROBLEM #62:  USE THE FORMULAS ON PAGE 400 (TEXT PAGE) FOR THIS ONE.  IT’S THE SAME PROCEDURE AS WITH LANE BUT THE FORMULA FOR THE STD DEV, HENCE Z IS SLIGHTLY DIFFERENT, BUT THE SAME TABLES ARE USED.

PROBLEM #70:  PAY ATTENTION TO ALL THE CHOICES THAT ARE CORRECT AS WELL.  REMEMBER THEM.

PROBLEM #96:  DO THE MATH, READ THE TABLES NOT SOFTWARE.  SIMPLY CALCULATE THE Z-SCORES FOR 85 AND 125 USING THE FORMULAS AND GET THE AREAS (PROBABILITIES/PERCENTAGES) FROM THE TABLES (THESE AREAS ARE VERY LARGE AND VERY SMALL).  SUBTRACT THE AREA FOR 85 FROM THAT FOR 125 TO GET THE AREA BETWEEN THEM.  THIS AREA IS THE PERCENTAGE OR PROBABILITY ASKED FOR.  (IT’S BIG)

 

Z-VALUES-N-DIST-WK4 (1).xlsx

MEAN
6.0000 WHEN WE HAVE LARGE DATA SETS, WE GROUP THE DATA. IN THIS CASE OUR GROUPS WILL BE:
STD DEV
2.17 IN A NORMAL DISTRIBUTION THE MEAN, MEDIAN AND MODE ARE ALL THE SAME NUMBER
X-VALUES Z-VALUES
GROUP NUMBER OF DATA POINTS IN IT
2
-1.84
2.0-2.9
2
2
-1.84
3.0-3.9
3
7
3
-1.38
4.0-4.9
4
6
3
-1.38
5.0-5.9
5
3
-1.38
6.0-6.9
6
5
4
-0.92
7.0-7.9
5
4
-0.92
8.0-8.9
4
4
4
-0.92
9.0-9.9
3
3
4
-0.92
10.0-10.9
2
5
-0.46
2
5
-0.46
5
-0.46
1
5
-0.46
5
-0.46
0
12
23
34
45
78
6
0.00
X-VALUES:
65
76
(3.83)
6
0.00 98 109 6
6
6
6 0.00
0.00
0.00
0.00 7 0.46 7
7
7 0.46
0.46
0.46 Z-VALUES ARE CALCULATED AND SHOW THE NUMBER OF STANDARD DEVIATIONS A DATA POINT (OR GROUP OF DATA POINTS) IS FROM THE MEAN
Z = ( X - MEAN) / STD DEV
OF COURSE IF YOU HAVE A Z-VALUE YOU CAN BACK CALCULATE TO GET THE X-VALUE: X = Z TIMES THE STD DEV + THE MEAN
AS AN EXAMPLE USE THE Z-VALUE OF +0.46 TO SEE WHAT X-VALUE IT REPRESENTS: X = 0.46 x 2.17 + 6.0 = 7.0 7
8
8
8
8
9
9
9 0.46
0.92
0.92
0.92
0.92
1.38
1.38
1.38 SIMILARLY THE X-VALUE (OR GROUP) THAT IS -1 STANDARD DEVIATION BELOW THE MEAN (Z = -1) WOULD BE: X = 1 x 2.17 + 6.0 = 3.83
WHILE THE SHAPE OF OUR EARLIER GRAPH WAS RECTANGULAR, THIS GRAPH HAS A BELL SHAPE, REFERRED TO AS THE "NORMAL" DISTRIBUTION.
AND, AS WITH THE RECTANGULAR GRAPH THE AREA SHOWN IN THIS GRAPH REPRESENTS 100 % OF OUR DATA AND ANY PORTION OF IT REPRESENTS
THE PERCENTAGE OF OUR DATA IN THAT AREA AS WELL AS THE PROBABILITY THAT SPECIFIC DATA ARE IN THAT REGION.
HERE IS A MORE TYPICAL BELL CURVE GRAPH:
TO CALCULATE THE AREA IN ANY SECTION OF THIS BELL CURVE GRAPH WE NEED TO KNOW THE EQUATION FOR THE LINE,
AND THEN WE USE INTEGRAL CALCULUS TO CALCULATE THE AREA UNDER THAT CURVE FOR THE AREA SELECTED.
n 10
10 1.84
1.84 Z-VALUES: -2 -1 0 +1 +2 THE X-VALUES ARE THE DATA POINTS AND WE HAVE GROUPED THEM AS SHOWN. ʃ
x= 1 FORTUNATELY FOR US, THESE AREAS HAVE BEEN CALCULATED AND PUT INTO TABLES. THE HORIZONAL AXIS HAS THE Z-VALUES ON IT, WHICH
ARE THE STANDARD DEVIATIONS FROM THE MEAN AND THE TABLE GIVES US THE AREA TO THE LEFT OF THAT Z-VALUE. (SEE & PRINT THE ATTACHED TABLES) TO GET THE AREA TO THE RIGHT WE MUST SUBTRACT THE TABLE VALUE FROM 1.0000 SINCE THE SUM OF BOTH AREAS EQUALS 100% OR 1.00 AS A DECIMAL.
HERE ARE SOME "WORKED-OUT" AREAS:
68.3% OF THE AREA (OUR DATA VALUES) ARE WITHIN -1 AND +1 STANDARD DEVIATIONS FROM THE MEAN
95.4% OF THE AREA (OUR DATA VALUES) ARE WITHIN -2 AND +2 STANDARD DEVIATIONS FROM THE MEAN
99.7% OF THE AREA (OUR DATA VALUES) ARE WITHIN -3 AND +3 STANDARD DEVIATIONS FROM THE MEAN
AN "UNUSUAL" VALUE IS CONSIDERED ONE THAT IS MORE THAN TWO STANDARD DEVAIONS FROM THE MEAN (Z = + 2) MORE SPECIFICALLY, THIS "UNUSAL" OR RARE EVENT AREA CORRESPONDS TO 5% OF THE GRAPH AREA
ACCORDINGLY, THESE CRITICAL Z-VALUES ARE +1.645 ON THE RIGHT END AND -1.645 ON THE LEFT END OF THE GRAPH.
SEE IF YOU CAN FIND THE AREA IN THE TABLE (0.05) FOR THESE Z-VALUES.

Attachments:

• 1492763699-Z-VALUES-N-DIST-WK4 1.xlsx