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LECTURE 5 OF 6 SUBTOPIC 12.2 CONTINUOUS PROBABILITY DISTRIBUTION 1 At the end of the lesson, students are able to: (a)Understand the normal distribution N , 2 (b) Find the mean and variance for the normal distribution. (c) Standardize the normal random variable and use the tables. 2 The Continuous Random Variables 40 30 20 10 160 165 170 175 180 185 190 The histogram will approach a well defined smooth curve called probability density function (p.d.f ) if the sample size of a continuous variable is increased and the class intervals are made as small 3 as possible. The Normal Distribution X N ( , ) 2 X continuous random va riable the mean value of X in the population the sta ndard deviation of X in the population 2 the va riance of X in the population Examples of continuous random variable : -The height of children of a certain age. - The diameters of metal cylinders. - The time taken by students to run 100 m. 4 The Normal Distribution X N ( , ) 2 X is a continuous random variable with p.d.f : a b 1 f ( x) e 2 ( x )2 2 2 , x The PROBABILITY is represented by the AREA of the shaded section of the graph. b P(a x b) f ( x) dx a 5 Properties of The Normal Distribution 1.The p.d.f curve is characterized by a symmetrical bell shaped curve and it is symmetric about the mean. So the mean, mode and median coincide. 2. The curve extends to infinity in either direction X , and 0 . 3. The total area under the curve is 1 . f ( x) dx 1 6 4. The maximum value for f(x) is approximately 1 when mean = mode = median= . 2 5. The probability that X lies in the interval from a to b is equal to the area under the curve bounded by the ordinates x = a and x = b . b P(a x b) f ( x) dx a 0a b 7 The Standard Normal Distribution Z N (0,1) where Z = the standard normal random variable , 2 mean( ) = 0 and variance( ) = 1 To find the probability of X ,continuous random variable ,we need to change the normal dist. into the standard normal dist. by using the transformation formula. 8 Transformation Formula Z X Z st andard normal random va riable X continuous random variable the mean value of X in the population the st andard deviation of X in the population 9 Transformation Formula Z X N ( , ) 2 P( X x) x X Z N (0,1) P( Z z ) 0 z 10 How to use the Standard Normal Tables Use the symmetrical property of the graph to find the probability from the tables. P( Z 0) 0.5 , P( Z 0) 0.5 Total area under the curve =1 z=0 11 Symmetrical Properties 1. P( Z a) P( Z a) Equal Area = -a 0 0 a Read the probability from the table with z=a. The standard normal table shows probability for 12 P ( Z z ) the ‘upper end’ area that is . 2. P ( Z a ) 1 P ( Z a ) 1 P( Z a) Equal Area = -a 0 P( Z a) 0 a P( Z a) 13 3. P(b Z a) P( a Z b ) P ( Z a ) P ( Z b) Equal Area = -b -a 0 P(b Z a) 0a b P ( a Z b) 14 4. P(a Z 0) P(0 Z a ) P( Z 0) P( Z a) 0.5 P( Z a) Equal Area = -a 0 0 a 15 Standard Normal Probability Tables The table shows 0.01positive 0.02 0.03 z 0.00 only values of Z.Use the 0.0 0.5000 0.4960 0.4920 0.4880 0.1 0.4602symmetrical 0.4562 0.4522 0.4483 0.2 0.4207 0.4168 0.4129 0.4090 property of the 0.3 0.3821 0.3783 0.3745 0.3707 graph . 0.3336 0.4 0.3446 0.3409 0.3372 0.04 0.05 0.06 0.07 0.08 0.09 0.4840 0.4443 0.4052 0.3669 0.3300 0.4801 0.4404 0.4013 0.3632 0.3264 0.4761 0.4364 0.3974 0.3594 0.3228 0.4721 0.4325 0.3936 0.3557 0.3192 0.4681 0.4286 0.3897 0.3520 0.3156 0.4641 0.4247 0.3859 0.3483 0.3121 i.e: P(Z > 0.05)= 0.4801 P( Z 0.48) P( Z 0.48) = 0.3156 The complete table shows the probability for values of z from 0.00 to 4.09. 16 For z > 4.09, P(Z > 4.09) is less than 0.00002 Example 1 If Z N(0,1) , find the probability of the following (a ) P( Z 2.5) ( b) P( Z 2.5) (c) P( Z 0.6 ) (d ) P( 2.23 Z 1.23) (e) P( Z 1.2) (f ) P( Z 1.78 ) 17 Solution (a) P( Z 2.5) 0.00621 0 2.5 (b) P( Z 2.5) 1 P Z 2.5 1 0.00621 0.99379 0 2.5 18 (c) P( Z 0.6) 1 P( Z 0.6) 1 P( Z 0.6) 1 0.2743 -0.6 0 0.7257 (d ) P(2.23 Z 1.23) P(1.23 Z 2.23) P( Z 1.23) P( Z 2.23) 0.1093 0.0129 0.0964 -2.23 -1.23 19 (e) P( Z 1.2) P( Z 1.2) P( Z 1.2) 2 P( Z 1.2) 2(0.1151) -1.2 0 1.2 0.2302 ( f ) P( Z 1.78) P(1.78 Z 1.78) 1 2 P( Z 1.78) 1 2(0.0375) 0.925 -1.78 0 1.78 20 Example 2 The random variable Z has a standard normal distribution. Determine the value of a if given: (i) P( Z a) 0.3745 (ii) P( Z a) 0.6217 (iii) P( Z a) 0.8472 (iv) P( Z a) 0.2714 21 Solution (i ) P( Z a) 0.3745 From the table :P ( Z 0.32) 0.3745 a 0.32 (ii) P( Z a) 0.6217 Since the probability > 0.5, so a < 0. Thus, 0.6217 1 - 0.6217=0.3783 a 0 22 P( Z a) 0.3783 From the table : P ( Z 0.31) 0.3783 a 0.31 (iii) P( Z a) 0.8472 P( a Z a) 0.8472 1 2 P( Z a) 0.8472 0.31 From the table : P( Z 1.43) 0.0764 a 1.43 1 0.8472 P( Z a) 0.0764 2 23 (iv) P( Z a) 0.2714 P(Z a) P Z a 0.2714 2 P( Z a) 0.2714 P( Z a) 0.1357 From the table : P( Z 1.1) 0.1357 a 1.1 24 Example 3 The random variable X has a normal distribution with mean 10 and standard deviation 2. Find : (a) P( X 13) (b) P 9 X 13 (c) the value of a if P (a X 12) 0.15 25 Solution X 2 N (10, 2 ) (a) P( X 13) X PZ Z X Given : 10 , 2 13 10 PZ 2 P( Z 1.5) 0.0668 26 (b) P 9 X 13 13 10 9 10 P Z 2 2 P( 0.5 Z 1.5) -0.5 0 1.5 1 P( Z 0.5) P Z 1.5 1 0.3085 0.0668 0.625 27 (c) the value of a if P(a X 12) 0.15 P( a X 12 ) 0.15 12 10 a 10 P Z 0.15 2 2 a 10 P Z 1 0.15 2 a 10 PZ P( Z 1) 0.15 2 a 10 PZ 0.15 0.1587 0.3087 2 28 a 10 PZ 0.3087 2 From the table : P( Z 0.5) 0.3085 a 10 0.5 2 a 2(0.5) 10 a 11 29 Example 4 The heights of matriculation students in KMPP is normally distributed with a mean of 170 cm and a standard deviation of 5 cm. (a)What is the probability that a student, selected at random, has a height between 174 cm and 178 cm? (b) Out of a group of 150 matriculation students, how many would be expected to have a height less than 164 cm? 30 Solution Z X=height of students in cm X 170 , 5 178 170 174 170 Z a P(174 X 178) P 5 5 P(0.8 Z 1.6) P( Z 0.8) P Z 1.6 0.2119 0.0548 0.1571 31 164 170 b P( X 164) P Z 5 P( Z 1.2) P( Z 1.2) 0.1151 Number of students expected to have a height less than 164 cm is 150 (0.1151) 17.265 17 students 32 33 Example 5 A machine is supposed to cut up logs into pieces that are each 2 m. However , the machine is an old one, and while the pieces that it produces do have an average length of 2 m, 10% of the pieces that it produces are less than 1.95 m. Assuming that the lengths produced are normally distributed, determine the proportion of the pieces that are longer than 2.10 m. 34 Solution X = the length of a log Given : X N (2.0, ) , P( X 1.95) 0.10 2 Fi nd P( X 2.10). 1.95 2.0 P( X 1.95) P Z 0.10 From the table : P( Z 1.28) 0.1003 1.95 2.0 1.28 35 1.95 2.0 1.28 0.0390 2.10 2.0 P( X 2.10) P Z 0.0390 P( Z 2.56) 0.00523 0.5% of the pieces of log have length greater than 2.10 metres. 36