Statistics Homework 4

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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
3	-1.38		4.0-4.9	4
3	-1.38		5.0-5.9	5
3	-1.38		6.0-6.9	6
4	-0.92		7.0-7.9	5
4	-0.92		8.0-8.9	4
4	-0.92		9.0-9.9	3
4	-0.92		10.0-10.9	2
5	-0.46
5	-0.46
5	-0.46
5	-0.46
5	-0.46
6	0.00
6	0.00
6	0.00
6	0.00
6	0.00
6	0.00		THE X-VALUES ARE THE DATA POINTS AND WE HAVE GROUPED THEM AS SHOWN.
7	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
7	0.46						Z = ( X - MEAN) / STD DEV
7	0.46		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
7	0.46		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	0.46		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
8	0.92		WHILE THE SHAPE OF OUR EARLIER GRAPH WAS RECTANGULAR, THIS GRAPH HAS A BELL SHAPE, REFERRED TO AS THE "NORMAL" DISTRIBUTION.
8	0.92		AND, AS WITH THE RECTANGULAR GRAPH THE AREA SHOWN IN THIS GRAPH REPRESENTS 100 % OF OUR DATA AND ANY PORTION OF IT REPRESENTS
8	0.92		THE PERCENTAGE OF OUR DATA IN THAT AREA AS WELL AS THE PROBABILITY THAT SPECIFIC DATA ARE IN THAT REGION.
8	0.92		HERE IS A MORE TYPICAL BELL CURVE GRAPH:
9	1.38									TO CALCULATE THE AREA IN ANY SECTION OF THIS BELL CURVE GRAPH WE NEED TO KNOW THE EQUATION FOR THE LINE,
9	1.38									AND THEN WE USE INTEGRAL CALCULUS TO CALCULATE THE AREA UNDER THAT CURVE FOR THE AREA SELECTED.
9	1.38									n
10	1.84									ʃ
10	1.84									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.

2 3 4 5 6 5 4 3 2

X-VALUES: 2 3 4 5 6 7 8 9 10 (3.83)

Z-VALUES: -2 -1 0 +1 +2

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