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QUESTION

# LET'S SEE IF ANY OF THE ABOVE 30 DATA POINTS WOULD BE CONSIDERED "UNUSUAL". HOW?

LET'S SEE IF ANY OF THE ABOVE 30 DATA POINTS WOULD BE CONSIDERED "UNUSUAL". HOW?

WE MUST DECIDE AT WHAT SIGNIFICANCE LEVEL WE WOULD CONSIDER A DATA POINT "UNUSUAL". IF WE CHOSE A SIGNIFICANCE LEVEL OF 10% THAT MEANS THAT A DATA POINT WOULD HAVE TO HAVE A POSITIVE Z-VALUE (STANDARD DEVIATION) THAT CORRESPONDED TO A TABLE AREA OF 0.9000 TO THE LEFT (SINCE THIS DATA POINT IS IN THE 10% AREA IN THE FAR RIGHT TAIL.   OR, IF THE DATA POINT HAD A NEGATIVE Z-VALUE ITS VALUE WOULD HAVE TO CORRESPOND TO A TABLE AREA OF 0.10 (10%) IN THE FAR LEFT TAIL OF THE CURVE.

A SIGNIFICANCE LEVEL OF 5% WOULD NEED A +Z VALUE CORRESPONDING TO A TABLE AREA OF 0.9500 TO THE LEFT (LEAVING 0.0500 TO THE FAR RIGHT). OR A -Z-VALUE CORRESPONDING TO A TABLE AREA OF SIMPLY 0.0500 TO THE LEFT IN THE FAR LEFT TAIL.

a) FILL IN THE BLANKS: A SIGNIFICANCE LEVEL OF 1% WOULD NEED A +Z VALUE CORRESPONDING TO A TABLE AREA OF ________ TO THE LEFT (LEAVING ________ TO THE FAR RIGHT). OR A -Z-VALUE CORRESPONDING TO A TABLE AREA OF SIMPLY _________ TO THE LEFT IN THE FAR LEFT TAIL

b)  IN QUESTION (2) YOU DETERMINED THE GENERAL CRITICAL Z-VALUES FOR SIGNIFICANCE LEVELS OF +10%, +5% AND +1%, SO COMPARE YOUR STANDARDIZED DATA TO THEM AND LIST YOUR Z-VALUES AND ORIGINAL X-VALUES THAT ARE "UNUSUAL" AT THESE SIGNIFICANCE LEVELS.    LIST ANY "UNUSUAL" VALUES. SO WHAT?

4) THE SPEED OF VEHICLES ALONG A STRETCH OF I-95 HAS AN APPROXIMATELY NORMAL DISTRIBUTION WITH A MEAN OF 77 MPH AND A STANDARD DEVIATION OF 9 MPH.

(a). THE SPEED LIMIT IS 70 MPH. WHAT IS THE PROPORTION OF VEHICLES GOING LESS THAN OR EQUAL TO THE SPEED LIMIT?

(b) WHAT PROPORTION OF THE VEHICLES WOULD BE GOING LESS THAN 65 MPH?

(c)  A NEW SPEED LIMIT WILL BE INITIATED SUCH THAT APPROXIMATELY 10% OF VEHICLES WILL BE OVER THAT SPEED LIMIT. WHAT IS THE NEW SPEED LIMIT BASED ON THIS CRITERION? (NEED TO CALCULATE THE X-VALUE)

(d)  DO YOU THINK THE ACTUAL DISTRIBUTION (HOW THE CURVE LOOKS) OF SPEEDS DIFFERS FROM A NORMAL BELL-SHAPED DISTRIBUTION?

5)  STUDENTS TAKE A STATISTICS TEST. THE GRADE DISTRIBUTION IS NORMAL WITH A MEAN OF 70, AND A STANDARD DEVIATION OF 6.

(a)  ANYONE WHO SCORES IN THE TOP 20% OF THE DISTRIBUTION GETS A GRADE OF "A" OR "B" WHAT IS THE LOWEST SCORE SOMEONE CAN GET AND STILL GET A "B"?

(b) THE BOTTOM 20% GET A "D" OR "F". WHAT IS THE LOWEST SCORE THAT STILL PASSES WITH THE "C" ?

6) WE CAN USE THE NORMAL DISTRIBUTION TO APPROXIMATE THE BINOMIAL DISTRIBUTION. YOU REMEMBER THE COMPLICATED BINOMIAL EQUATIONS (WEEK 3)? THE EQUATIONS USING THE NORMAL TO APPROXIMATE THE BINOMIAL ARE MUCH SIMPLER, BUT THEY ARE NOT AS PRECISE. LET'S SOLVE THIS SAME BINOMIAL PROBLEM USING THE NORMAL DISTRIBUTION SHORTCUT AND SEE HOW CLOSE IT IS TO OUR BINOMIAL CALCULATION OF WEEK 3.

Here is a web site that offers a decent explanation of what we will be doing here: http://onlinestatbook.com/2/normal_distribution/normal_approx.html

You can also check LANE around page 204 AND 264 for a review of the DISCRETE data handling.

Problem Statement from Week 3: A coin is weighted so that heads comes up 75% (0.75) of the time and tails 25%. What is the probability you get 4 or more heads in 5 tosses of that weighted coin?

What is the probability you determined this to be using the complex equation from Week 3? P(4) + P(5) = _____

Now to use the Normal Distribution as an approximation, we first need to calculate the MEAN and STANDARD DEVIATION of that data set.  In addition to the above web site, our LANE text shows these formulas around page 204. We needs these statistics to STANDARDIZE our normalized discrete data as you will see below.

The MEAN = number of trials (data points), in this case 5 tosses times the probability of HEADS, where here it's 75% or 0.75. The MEAN is therefore: 5 x 0.75 =________

The VARIANCE is the number of trials TIMES the probability of winning (HEADS) then TIMES the probability of losing (TAILS) (losing is simply 1.00 - probability of winning since wins and losses must have a total probability of 100% or 1.00 so no "ties" allowed): The VARIANCE is therefore: 5 x 0.75 x (1.0 - 0.75) = _______ The STANDARD DEVIATION is simply the SQUARE ROOT of the VARIANCE:  so it is: _______

Back to the COIN PROBLEM. The number of trials is N = 5. The number of HEADS (here we want 4 AND 5) are DISCRETE numbers representing wins and losses, so getting 1.345 heads is NOT possible. If we graphed discrete numbers, like the number of heads, it would be a bar chart (NOT a histogram).

However, the NORMAL DISTRIBUTION represents CONTINOUS DATA where fractions are possible. So, how do we handle discrete data? Kind of simple, take the discrete value of 5 heads. To make it continuous it would go from 4.5 to 5.5 continuous wins. Four heads would be from 3.5 to 4.5. (FYI: ZERO heads would be a little trickier. It would be represented by a continuous range of -0.5 to +.05 )

Let's proceed to calculating the probability using this NORMAL APPROXIMATION. For the desired outcome of 4 or 5 heads we have the continuous range for our discrete data of 3.5 to 5.5  Consider these ends a x-values (3.5 and 5.5) NEXT we STANDARDIZE the two ends of this range using the MEAN and STANDARD DEVIATION CALCULATED ABOVE:

The x-value of +3.5 becomes the standardized z-value of:_____________ and +5.5 becomes the z-value of: ________

(KEEP TWO DECIMAL PLACES LIKE 1.45 OR 0.98)

These z values are the standard deviations (SD) from the mean that we go to our TABLE with to get the probabilities of tossing 4 heads (the LOWER SD) and 5 heads (UPPER SD) out of the 5 weighted coin tosses.

The upper SD probability is _________ meaning that this percent of our data have that probability (or less) of occurring. In this case it is getting 5 heads (or fewer since it is the entire area to the left) out of 5 tosses. The lower SD probability is ________, which is the probability of 4 (or fewer) heads out of 5 tosses.

SINCE THESE ARE AREAS TO THE LEFT WE MUST SUBTRACT THE LOWER AREA (PERCENTAGE OR PROBABILITY) FROM THE UPPER AREA. THE RESULT IS THE PROBABILITY OF GETTING 4 OR 5 HEADS OUT OF 5 TOSSES. WHAT DID YOU CALCULATE? ______________  HOW CLOSE IS THIS NORMAL APPROXIMATION TO THE MORE ACCURATE DISCRETE NUMBER YOU CALCULATED IN WEEK 3?

8) HEIGHT AND WEIGHT ARE TWO MEASUREMENTS USED TO TRACK A CHILD'S DEVELOPMENT. THE WEIGHTS FOR ALL 11 YEAR OLD GIRLS, 4' 8" TALL IN A REFERENCE POPULATION HAD A MEAN  OF µ = 73 POUNDS  WITH A STANDARD DEVIATION OF σ = 1.5 LBS. ASSUME THESE WEIGHTS ARE NORMALLY DISTRIBUTED. CALCULATE THE Z-SCORES THAT CORRESPOND TO THE FOLLOWING WEIGHTS AND INTERPRET THEM. THIS IS USEFUL STATISTICS.

(a)  70 LBS

(b)  84 LBS

(c)  61 LBS

(d) IF YOU WERE THE PARENT OF ANY OF THESE CHILDREN, WOULD YOU BE CONCERNED? (WHY?)

9) A HYPOTHETICAL LEGAL STATISTICAL PROBLEM: A PATERNITY LAWSUIT. THE LENGTH OF A PREGNANCY IS NORMALLY DISTRIBUTED WITH A MEAN OF 285 DAYS AND A STANDARD DEVIATION OF 15 DAYS. AN ALLEGED FATHER WAS OUT OF THE COUNTRY FROM 235 TO 300 DAYS BEFORE THE BIRTH OF THE CHILD, SO THE PREGNANCY WOULD HAVE BEEN LESS THAN 235 DAYS OR MORE THAN 300 DAYS LONG IF HE WAS THE FATHER.  A HEALTHY CHILD WAS BORN WITH NO COMPLICATIONS, BUT:

(a) WHAT IS THE PROBABILITY THAT HE IS NOT THE FATHER?

(b) WHAT IS THE PROBABILITY THAT HE COULD BE THE FATHER?

(HINT: CALCULATE THE Z-SCORES FIRST, AND THEN USE THOSE TO DETERMINE THE PROBABILITIES)

10) (A NEW WRINKLE IN THIS PROBLEM SINCE IT DEALS WITH AVERAGES AND NOT SPECIFIC DATA VALUES. CHECK ILLOWSKY CHAPTER 7 SECTION 7.1. THE X WITH A BAR OVER IT REPRESENTS AN AVERAGE - I REFER TO IT AS X-BAR)

SUPPOSE THAT THE DISTANCES OF FLY BALLS HIT TO THE OUTFIELD (IN BASEBALL) IS NORMALLY DISTRIBUTED WITH A MEAN OF 235 FEET AND A STANDARD DEVIATION OF 35 FEET. WE RANDOMLY SAMPLE 50 FLY BALLS. IF  = AVERAGE DISTANCE IN FEET FOR 50 FLY BALLS, THEN

(a)  WHAT IS THE PROBABILITY THAT THE 50 FLY BALLS TRAVELED AN AVERAGE OF LESS THAN 225 FEET? SKETCH THE GRAPH. SCALE THE HORIZONTAL AXIS FOR  . SHADE THE REGION CORRESPONDING TO THE PROBABILITY. FIND THE PROBABILITY.

(b) FIND THE 70TH PERCENTILE OF THE DISTRIBUTION OF THE AVERAGE OF 50 FLY BALLS