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Quick review of some key ideas CEE 11 Spring 2002 Dr. Amelia Regan These notes draw liberally from the class text, Probability and Statistics for Engineering and the Sciences by Jay L. Devore, Duxbury 1995 (4th edition) mean The mean of a sample or data set is simply the arithmetic average of the values in the set, obtained by summing the values and dividing by the number of values. x1 x2 ...xn 1 n x xi n n i 1 The mean of the sample of weights is 144.63 pounds mean of a frequency distribution When we summarize a data set in a frequency distribution, we are approximating the data set by "rounding" each value in a given class to the class mark. n 1 n x fi xi pi xi n i 1 i 1 where fi the frequency of the ith observation and pi = the proportion associated with the ith observation The mean of the weight data obtained in this way is 146.67 shortcut formula for the variance Its sometimes more convenient to use the following formula for the variance xi n 2 i 1 x i n i 1 n 1 n n s2 x x i 1 i n 1 2 2 properties of S2 Let x1, x2, x,...,xn be a sample and c be any nonzero constant. If y1 = x1 + c, y2 = x2 + c,...,yn = xn + c, then S2y = S2x If y1 = cx1, y2 = cx2,...,yn = cxn, then S2y = c2S2x, Sy = |c|S2x In other words -- if we add a constant to a sample we do not increase the variance -- if we multiply by a constant we increase the variance by the square of the constant related properties of the sample mean Let x1, x2, x,...,xn be a sample and c be any nonzero constant. If y1 = x1 + c, y2 = x2 + c,...,yn = xn + c then y x c If y1 = cx1, y2 = cx2,...,yn = cxn, then y cx In other words if we add or multiply the sample by a constant we add or multiply the mean by the same constant Class exercise (new) Without using your calculators, calculate the mean and variance of the following data Xi | 33 27 18 36 21 ---------------------------------------------fi | 4 3 2 5 1 Hint, shift the observations “to the left” by subtracting a constant and then divide by another constant Class exercise (new) Without using your calculators, calculate the mean and variance of the following data Step 1 yi | 11 9 6 12 ---------------------------------------------Step 2 yi | 2 0 -3 3 yi | 2 0 -3 3 -2 ---------------------------------------------fi | 4 3 2 5 1 Divide by 3, then subtract 9 7 -2 xi yi 9 3 yi | 2 0 -3 3 -2 -------------------------------------------------------(yi)2 | 4 0 9 9 4 -------------------------------------------------------fi | 4 3 2 5 1 f y 8 0 6 15 2 15 f y 15 y 1.0 i i i n i 15 fi y 2i 16 0 18 45 4 83 xi yi 9 3 2 fi yi 2 2 i 1 15 fi yi 83 n 15 4.847 n 1 14 n n i 1 xi 3( yi 9) x 3( y 9) 30.0 s 9s 43.714 2 x 2 y conditional probability For any two events A and B with P(B) > 0, the conditional probability of A given that B has occurred is defined by P( A B) P( A | B) P( B) The multiplication rule for P(A intersection B) follows directly P( A B) P( A | B) P( B) independence Remember that in general P( A B) P( A | B) P( B) However, if A and B are independent then P( A B) P( A) P( B) In fact, A and B are independent if and only if the above is true Counting Techniques When the various outcomes of an experiment are equally likely then the task of computing probabilities reduces to counting. In particular, if N is the number of outcomes in the sample space and N(A) is the number of outcomes contained in an event A, then N ( A) P ( A) N Permutations Any ordered sequence of k objects taken from a set of n distinct objects is called a permutation of size k of the objects. The number of permutations of size k that can be constructed from the n objects is denoted by Pk,n The number of permutations of size k that can be constructed from n objects is equal to n(n-1)(n-2)…(nk+1) Pk , n n(n 1),...( n k 1) n(n 1),..., ( n k 1)( n k )( n k 1) ( n k )( n k 1),...(2)(1) n! ( n k )! combinations Given a set of n distinct objects, any unordered subset of size k of the objects is called a combination. The number of combinations of size k that can be formed from n distinct objects is denoted by nk or sometimes by Ck,n n Pk ,n n! k! k !(n k )! k