Normal Dist

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NORMAL DISTRIBUTION

IS AN APPROXIMATION TO BINIOMIAL DISTRIBUTION WHEN - NUMBER OF TRIALS n IS VERY LARGE - p AND q ARE NOT VERY SMALL IS DEFINED FOR CONTINUOUS TYPE OF VARIABLES NOTE: BINOMIAL DISTRIBUTION TENDS TO THE FORM OF THE CONTINUOUS CURVE WHEN n BECOMES LARGE.

IS A LIMITING CASE OF POISSON DISTRIBUTION WHEN ITS MEAN IS LARGE

IT IS AN EXTENSION OF BOTH BINOMIAL AND POISSON DISTRIBUTION WHEN n IS LARGE

GRAPH OF NORMAL DISTRIBUTION / PROPERTIES OF NORMAL CURVE

1) BELL SHAPED SYMMETRICAL CURVE, SYMMETRY ABOUT MEAN.

2) TAILS NEVER TOUCH X AXIS AND EXTENDS TO INFINITY IN BOTH DIRECTIONS

3) THE CURVE IS UNIQUE FOR A PARTICULAR PAIR OF MEAN AND STANDARD DEVIATION.

4) MEAN, MEDIAN, MODE ARE ALL EQUAL AND COINCIDE AT THE CENTR OF THE CURVE ABOUT WHICH THE CURVE IS SYMMETRIC.

5) AREA UNDER THE NORMAL CURVE IS DISTRIBUTED AS FOLLOWS a) MEAN 1 b) MEAN 2 c) MEAN 3 COVERS 68.27 % AREA COVERS 95.45 % AREA COVERS 99.73 % AREA

THE NORMAL DISTRIBUTION CURVE

f( )

68.27% 95.45% 3 - -2 - MEAN 99.73% 2 3

z

PROBABILITIES AND AREAS UNDER NORMAL CURVE

THE PROBABILITY THAT A NORMAL VARIABLE WILL BE WITHIN 1 STANDARD DEVIATION FROM ITS MEAN (ON EITHER SIDE) IS 0.6827 THE PROBABILITY THAT A NORMAL VARIABLE WILL BE WITHIN 2 STANDARD DEVIATIONS FROM ITS MEAN IS 0.9545 THE PROBABILITY THAT A NORMAL VARIABLE WILL BE WITHIN 3 STANDARD DEVIATION FROM ITS MEAN IS 0.9973

STANDARD NORMAL PROBABILITY DISTRIBUTION PORTIONS OF THE AREA UNDER THE NORMAL CURVE THAT ARE CONTAINED WITHIN ANY NUMBER OF STANDARD DEVIATIONS FROM THE MEAN CAN BE EXTRACTED FROM STATISTICAL TABLES FOR EACH PAIR OF MEAN AND S.D THERE EXISTS AN UNIQUE CURVE AND THUS A NEED OF UNIQUE STATISTICAL TABLE (NOT POSSIBLE)

CONVERT ANY GIVEN NORMAL DISTRIBUTION TO STANDARD NORMAL DISTRIBUTION(MEAN=0; S.D =1

CONVERSION PROCESS.. TRANSFORMATION RULE:

Z=

X X

VARIABLE Z MEASURES DEVIATION FROM MEAN IN UNITS OF STANDARD DEVIATION. Z IS CALLED STANDARDISED VARIABLE AND ITS VALUE IS CALLED STANDARD SCORE.

STANDARD SCORE GIVES US THE NUMBER OF STANDARD DEVIATIONS A PARTICULAR LIES BELOW OR ABOVE THE MEANX +1.8

Z=1.8

IMPLIES

X=

MEANING X IS 1.8 TO THE RIGHT OF MEAN IN X- SCALE

NORMAL CURVE 68.27% 95.45% 3 STANDARD NORMAL CURVE 2 X

MEAN = S.D =

X

2 99.73%

3

68.27%

MEAN=0 S.D=1 3

-3

-2

95.45% 1 Z=0 1 2 99.73%

X

GIVEN MEAN=82, S.D =1.5

P[80 .8 < X < 83 .2] = P[(80 .8 82 ) / 1.5 < Z < (83 .2 82 ) / 1.5] = P[.8 < Z < .8] = .7881 .2119 = .5762

TABLE OF Z- TRANSFORM GIVES AREA UNDER CURVE FOR +VE SIDE ONLY. [BETWEEN 0 TO Z]. FOR AREA ON VE SIDE, FIND AREA ON +VE SIDE FOR THE SAME VALUE (SYMMETRIC CURVE) P(0 Z 1.20)= P(-1.20 Z 0)

0

z

FIND AREA UNDER THE NORMAL CURVE IN EACH OF THE FOLLOWING CASES1) P( 0 Z 1.2)

ANS: .3849

0 1.2

2) P( -.68 Z 0) = P( 0 Z .68)= .2518 3) P( -.46 Z 2.21)

-.46

2.21 P( -.46 Z 0) +P( 0 Z 2.21) = .1772 + .4864 = .6636

4) P( .81 Z 1.94 )

.81

1.94

P( 0 Z 1.94) P( 0 Z .81) =.4738 - .2910 = .1828

5) TO THE LEFT OF Z = - .6 =.5 - P(0 Z .6) ANS: .2743 -.6 6) TO THE RIGHT OF Z= -1.28=.5 + P(0 Z 1.28)

ANS: .8997 -1.28

7) RIGHT OF Z= 2.05 AND LEFT OF Z= -1.44

-1.44=1- P(-1.44 Z 2.05)

2.05

= 1-[P(0 Z 1.44)+ P(0 Z 2.05)]= .0951

Q. A NORMAL CURVE HAS MEAN AS 20 AND S.D AS 10 FIND AREA BETWEEN 15 AND 40. HINT: P(15 X 40) SOL: .6687 Q. ASSUME MEAN HEIGHT OF SOLDIERS TO BE 68.22 INCHES WITH VARIANCE OF 10.8 INCHES. HOW MANY SOLDIERS IN A REGIMENT OF 1,000 ARE EXPECTED TO BE OVER SIX FEET TALL. HINT: 6FEET = 72 INCHES=X, FIND P(X 72 )x1000 SOL: 125

Q. IN A TEST GIVEN TO 1,000 STUDENTS, THE AVERAGE SCORE WAS 42 AND SD = 24. FIND A) NUMBER OF STUDENTS EXCEEDIND A SCORE OF 50. B) NUMBER OF STUDENTS BETWEEN 30 AND 54 SOL: A) 1000x .3696 B) 383

Q. SUPPOSE THAT THE ACTUAL AMOUNT OF INSTANT COFFEE WHICH A FILLING MACHINE PUTS INTO 5 OUNCE JARS IS A FOLLOWING A NORMAL DISTRIBUTION WITH S.D= .04 OUNCE. IF ONLY 3% OF THE JARS ARE TO CONTAIN LESS THAN 5 OUNCES OF COFFEE,WHAT MUST BE THE MEAN FILL OF THESE JARS?

SOL: =.04 P(X 5)=.03

.47

Z=5X

X X

FIND Z S.T AREA B/W O AND Z IS .47 .47 .03

5 X Z = = .88 1 .04 X =1.88 .04 +5 =5.0752

Z 0

Q. ASSUME IN A DISTRIBUTION EXACTLY NORMAL, 7% OF THE ITEMS ARE UNDER 35 AND 89% ARE UNDER 63. WHAT IS THE MEAN AND S.D OF THE DISTRIBUTION ? SOL:Z CORRESPONDING TO X=35 AND AREA .43 ON LEFT IS 1.48 Z CORRESPONDING TO X=63 AND AREA .39 ON RIGHT IS 1.23 MEAN=50.3, S.D=10.33

7% 35 89%X

63 39%

7% Z1 Z=0 Z243%

Q. OF A LARGE GROUP OF MEN 5% ARE UNDER 60 INCHES IN HEIGHT AND 40% ARE BETWEEN 60 & 65 INCHES. ASSUMING A NORMAL DISTRIBUTION, FIND MEAN HEIGHT AND S.D. ANS: MEAN=65.429; S.D= 3.3

CASE

IN A MANUFACTURING UNIT STEEL RODS ARE MANUFACTURED TO BE 3 INCHES IN DIAMETER , BUT ARE ACCEPATABLE IF THEY ARE WITHIN THE ERROR OF .01. IT IS OBSERVED THAT 15% ITEMS ARE OVER DIMENSIONED AND 15 % OF THE ITEMS ARE UNDER DIMENSIONED. FIND THE MEAN AND STANDARD DEVIATION OF THE DISTRIBUTION.

HINT: 15% OVER DIMENSIONED P( X 3.01)= .15 15% UNDER DIMENSIONED P(X 2.99)= .15 SOLUTION: MEAN=3 S.D=.00962

NORMAL APPROXIMATIONQ. USE NORMAL APPROXIMATION TO THE BINOMIAL DISTRIBUTION TO FIND THE PROBABILITY OF WINNING AT MOST 70 OF 100 MATCHES BY A TEAM, WHEN THE PROBABILITY OF WINNING EACH MATCH IS .75 HINT: p= .75, n=100, MEAN=np=.75*100=75, S.D= SQRT(npq) = 4.33 CONVERT DISRETE TO CONTINUOUS- APPLY CONTINUITY CORRECTION PROB OF WINNING NOT MORE THAN 70 MATCHES P(X 70.5)=P(Z -1.04)= .15

Q. FIND THE PROBABILITY OF GETTING 5 HEADS AND 10 TAILS IN 15 TOSSES OF A FAIR COIN USING NORMAL DISTRIBUTION. ALSO COMPARE YOUR ANSWER WITH BINOMIAL DISTRIBUTION . SOL: p BE THE PROB OF GETTING HEAD IN SINGLE TOSS =1/2 n=15 ; np=7.5 ; S.D= SQRT(npq)=1.94 NORMAL APPROX: P( X =5) P(4.5 X 5.5) = P(-1.55 Z -1.03)= .0909 BINOMIAL DISTRIBUTION: P(X=5)=n

c5 p 5 q n 5

= .09164 (APPROX. SAME)

CASE

Dr. Arnold of Lifeline Hospital is in charge of Research department. A new medicine is launched in the market and is known to be accompanied by bad side effects in 25% of the patients suffering from high blood pressure. Lifeline Hospital wants to carry out its own Research before actually prescribing the drug to the patients. For this purpose 120 patients with high blood pressure were treated with this new medicine.The drug would be rejected if probability of more than 32 patients suffering from side effects exceeds .25. Help Arnold in taking the decision.

SOLUTION p =.25 ; n =120 ; MEAN = np =30; S.D= 4.75 AFTER CONTINUITY CORRECTION P( X > 32) EQUIVALENT TO P(X 32.5) = P(Z .53)=.2019