- WELCOME TO A LESSON ON RANGE AND STANDARD DEVIATION.
IN THIS LESSON WE'LL DEFINE RANGE,
DEFINE STANDARD DEVIATION,
AND ALSO FIND THE RANGE AND STANDARD DEVIATION
OF A SET OF DATA.
THE RANGE IS THE DIFFERENCE BETWEEN THE MAXIMUM VALUE
AND THE MINIMUM VALUE OF THE DATA SET.
SO HERE WE HAVE A DATA SET THAT GIVES US THE AVERAGE GAS MILEAGE
OF 10 VEHICLES.
FIRST, NOTICE HOW THE DATA IS ORDERED FROM LEAST TO GREATEST,
WHICH MAKES IT EASIER TO IDENTIFY THE MAXIMUM VALUE
AND A MINIMUM VALUE.
NOTICE THE MAXIMUM VALUE IS 38, AND THE MINIMUM VALUE IS 12.
AND, THEREFORE, IF WE WANT TO FIND THE RANGE,
THE RANGE WILL BE THE DIFFERENCE BETWEEN THESE VALUES.
SO THE RANGE = 38 - 12 GIVING A RANGE OF 26.
NOW, LET'S TALK ABOUT STANDARD DEVIATION.
THE STANDARD DEVIATION IS A MEASURE OF VARIATION
BASED ON MEASURING HOW FAR EACH DATA VALUE DEVIATES
OR IS DIFFERENT FROM THE MEAN.
AND HERE ARE A FEW IMPORTANT CHARACTERISTICS
ABOUT STANDARD DEVIATION.
STANDARD DEVIATION IS ALWAYS POSITIVE.
STANDARD DEVIATION WILL BE ZERO IF ALL THE DATA VALUES ARE EQUAL
AND WILL GET LARGER AS THE DATA SPREADS OUT.
STANDARD DEVIATION HAS THE SAME UNITS AS THE ORIGINAL DATA.
AND STANDARD DEVIATION, LIKE THE MEAN,
CAN BE HIGHLY INFLUENCED BY OUTLIERS
OR VALUES THAT ARE VERY LARGE OR VERY SMALL
COMPARED TO THE REST OF THE DATA.
BEFORE WE DISCUSS HOW TO FIND THE STANDARD DEVIATION,
LET'S REVIEW HOW TO FIND THE MEAN OF A DATA SET.
THE MEAN OF A SET OF DATA IS THE SUM OF THE DATA VALUES
DIVIDED BY THE NUMBER OF VALUES.
REMEMBER, MEAN IS THE SAME AS THE AVERAGE.
SO IF WE WANT TO FIND THE MEAN OF THE DATA SET
FROM THE PREVIOUS SLIDE,
WHICH IS THE GAS MILEAGE OF 10 VEHICLES,
WE WOULD FIND THE SUM AND THEN DIVIDE BY 10
BECAUSE THERE ARE 10 DATA VALUES.
NOTICE HERE THE MEAN OR AVERAGE INDICATED BY X-BAR = 23,
AND THIS WOULD BE 23 MILES PER GALLON.
AND HERE'S HOW WE FIND THE STANDARD DEVIATION.
STEP ONE, WE FIND THE DEVIATION OF EACH DATA VALUE
FROM THE MEAN.
IN OTHER WORDS, WE SUBTRACT THE MEAN FROM EACH DATA VALUE.
WE INDICATE THIS BY X SUB N - X-BAR,
WHERE X SUB N WOULD BE ONE OF THE DATA VALUES
AND X-BAR IS THE MEAN.
STEP TWO, WE SQUARE THE DEVIATION,
OR SQUARE THE DIFFERENCES.
STEP THREE, WE ADD THE SQUARE DEVIATIONS.
STEP FOUR, WE DIVIDE BY N, THE NUMBER OF DATA VALUES,
IF THE DATA REPRESENTS A WHOLE POPULATION,
AND WE DIVIDE BY N - 1 IF THE DATA IS FROM A SAMPLE.
AND THEN FINALLY, STEP FIVE,
WE COMPUTE THE SQUARE ROOT OF THE RESULT.
AND HERE'S A FORMULA FOR STANDARD DEVIATION.
NOTICE HOW HERE WE'RE DIVIDING BY N - 1,
SO THIS WOULD BE THE FORMULA FOR A SAMPLE.
IF THIS WAS JUST N, IT WOULD BE THE FORMULA FOR A POPULATION.
SO, AGAIN, LOOKING AT THE NUMERATOR,
NOTICE HOW WE FIRST FIND THE DIFFERENCE OF EACH DATA VALUE
IN THE MEAN.
THEN WE SQUARE EACH DIFFERENCE.
THEN WE FIND THE SUM.
THEN WE DIVIDED BY EITHER N OR N - 1,
AND THEN, FINALLY, TAKE THE SQUARE ROOT.
SO IF WE WANT TO FIND THE STANDARD DEVIATION
OF OUR DATA SET--
AND, AGAIN, THE FIRST STEP IS TO FIND THE MEAN,
WHICH WE ALREADY DID, WHICH IS EQUAL TO 23.
NOW, THIS PROBLEM DOES NOT INDICATE
WHETHER THIS IS A SAMPLE OR POPULATION,
SO WE'LL ACTUALLY FIND BOTH STANDARD DEVIATIONS
WHERE THIS LETTER HERE IS CALLED SIGMA.
SO SIGMA SUB S WOULD BE THE SAMPLE STANDARD DEVIATION
AND SIGMA SUB P WOULD BE THE POPULATION STANDARD DEVIATION.
TO KEEP THIS NICE AND NEAT
I THINK IT IS IMPORTANT TO ORGANIZE THE INFORMATION
IN A TABLE, LIKE I'VE DONE HERE.
NOTICE HOW I LISTED THE DATA VALUES HERE IN THE FIRST COLUMN.
THE SECOND COLUMN TAKES EACH DATA VALUE
AND SUBTRACTS THE MEAN,
WHICH WOULD BE X SUB N - X-BAR.
SO 12 - 23, 14 - 23, AND SO ON.
SO, AGAIN, THIS IS THE FIRST STEP
WHERE WE FIND THE DIFFERENCE OF THE DATA VALUE AND THE MEAN.
NEXT, WE SQUARE THESE VALUES,
WHICH WE SEE HERE IN THE THIRD COLUMN.
AND FOR THE NEXT STEP WE'RE GOING TO SUM ALL THESE VALUES,
WHICH WOULD GIVE US THE NUMERATOR HERE
UNDERNEATH THE SQUARE ROOT.
SO NOTICE THE SUM OF THE DEVIATION SQUARED = 550,
WHICH, AGAIN, WOULD BE THIS VALUE HERE
UNDER THE SQUARE ROOT.
AND THEN BECAUSE WE HAVE 10 DATA VALUES, N = 10.
SO NOTICE FOR THE SAMPLE STANDARD DEVIATION,
N - 1 WOULD BE 9,
AND FOR THE POPULATION STANDARD DEVIATION, WE JUST DIVIDE BY N,
WHICH IS 10.
SO THE SAMPLE STANDARD DEVIATION WOULD BE THE SQUARE ROOT
OF 550 DIVIDED BY 9
WHICH IS APPROXIMATELY 7.8.
THE VALUE BEFORE WE TAKE THE SQUARE ROOT
IS CALLED THE SAMPLE VARIANCE.
AND FOR THE POPULATION STANDARD DEVIATION, 550 DIVIDED BY 10,
OR 55,
WOULD BE THE POPULATION VARIANCE.
SO NOTICE HOW THE POPULATION STANDARD DEVIATION
IS APPROXIMATELY 7.4,
AND THE SAMPLE STANDARD DEVIATION IS APPROXIMATELY 7.8.
AND, AGAIN, THE UNITS ON THESE WOULD BE MILES PER GALLON,
THE SAME AS THE UNITS IN THE ORIGINAL DATA.
NOW LETS FINISH BY TALKING ABOUT INTERPRETING
THE STANDARD DEVIATION.
THE STANDARD DEVIATION,
REPRESENTED BY THE GREEK LETTER SIGMA,
SHOWS HOW MUCH VARIATION FROM THE AVERAGE OR MEAN EXISTS.
A LOW STANDARD DEVIATION INDICATES THAT THE DATA VALUES
TEND TO BE VERY CLOSE TO THE MEAN.
A HIGH STANDARD DEVIATION INDICATES
THAT THE DATA POINTS ARE SPREAD OUT
OVER A LARGE RANGE OF VALUES.
THE STANDARD DEVIATION IS ALSO HELPFUL
WHEN COMPARING TWO SETS OF RELATED DATA.
CONSIDER THE RATING OF TWO AUTO REPAIR CENTERS.
COMPANY "A" AND COMPANY B HAVE MEAN RATINGS OF 3.0 STARS.
BUT COMPANY "As" RATINGS HAVE A STANDARD DEVIATION OF 0.5,
AND COMPANY Bs RATINGS HAVE A STANDARD DEVIATION OF 1.5.
SO THE QUESTION IS WHAT DOES THIS TELL US?
WELL, IT TELLS US THAT COMPANY Bs RATINGS ARE MORE SPREAD OUT
FROM THE MEAN RATING OF THREE STARS THAN COMPANY "As" RATINGS.
COMPANY "As" RATINGS HAS A SMALLER STANDARD DEVIATION,
SO MORE OF THE RATINGS ARE CLOSE TO THE MEAN OF THREE STARS.
WHICH MEANS FOR COMPANY B,
MORE PEOPLE GAVE HIGHER RATINGS AND LOWER RATINGS
THAN THREE STARS,
AND MORE OF COMPANY "A" RATINGS
WERE CLOSE TO THE AVERAGE OF THREE STARS.
I HOPE YOU HAVE FOUND THIS LESSON HELPFUL.