3. Continuous Random Variables

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(e) - Using interpolation: (f) Using tables "in reverse", . Statistics and probability: 3-8 ( ) (g) Finding a range of values within which lies with probability 0.95: The answer is not unique; but suppose we want an interval which is symmetric about zero i.e. between -d and d. Tail area = 0.025 � P(Z � d) = �(d) = 0.975 Using the tables "in reverse", d = 1.96. � range is -1.96 to 1.96. Example: Manufacturing variability The outside diameter, X mm, of a copper pipe is N(15.00, 0.02 2 ) and the fittings for joining the pipe have inside diameter Y mm, where Y ~ N(15.07, 0.022 2 ). (i) Find the probability that X exceeds 14.99 mm. (ii) Within what range will X lie with probability 0.95? (iii) Find the probability that a randomly chosen pipe fits into a randomly chosen fitting (i.e. X < Y). Solution (i) (ii) From previous example i.e. ( ( ) ) P=0.025 lies in (-1.96, 1.96) with probability 0.95. P=0.025
i.e. the required range is 14.96mm to 15.04mm. Statistics and probability: 3-9 (iii) For we want ). To answer this we need to know the distribution of Distribution of the sum of Normal variates Remember than means and variances of independent random variables just add. So if are independent and each have a normal distribution , we can easily calculate the mean and variance of the sum. A special property of the Normal distribution is that the distribution of the sum of Normal variates is also a Normal distribution. So if are constants then: 2 2 2 2 2 2 c X � c X �� c X ~ N( c � ��c � , c � � c � �� c � ) 1 1 2 2 n n 1 1 n n 1 1 2 2 n n Proof that the distribution of the sum is Normal is beyond scope. Useful special cases for two variables are If all the X's have the same distribution i.e. � 1 = � 2 = ... = � n = � , say and � 1 2 = �2 2 = ... = � n 2 = � 2 , say, then: (iii) All c i = 1: X 1 + X 2 + ... + X n ~ N(n�, n� 2 ) (iv) All c i = 1/n: X = X1 � X 2 �� X n n ~ N(�, � 2 /n) The last result tells you that if you average n identical independent noisy measurements, the error decreases by √ . (variance goes down as ). Example: Manufacturing variability (iii) Find the probability that a randomly chosen pipe fits into a randomly chosen fitting (i.e. X < Y). Using the above results Hence ( √ )
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3. Continuous Random Variables

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