hw7.solu

hw7.solu - STA 13A HW7 Solutions 6.6 a P ( z < 2.33= A (...

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6.6 a ( ) ( ) 2.33 2.33 .9901 Pz A <= = b As in part a , ( ) ( ) 1.645 1.645 . A However, the value for 1.645 z = is not given in Table 3, but falls halfway between two tabulated values, 1.64 z = and 1.65 z = . One solution is to choose an area ( ) 1.645 A which lies halfway between the two tabulated areas, ( ) 1.64 .9495 A = and ( ) 1.65 .9505. A = Then ( ) 1.645 .9500 A = and ( ) ( ) 1.645 .9500 . A This method of evaluation is called “linear interpolation” and is performed as follows: 1 The difference between two entries in the table is called a “tabular difference”. Interpolation is accomplished by taking appropriate portions of this difference. 2 Let P 0 be the probability associated with z 0 (i.e. ( ) 00 P Az = ) and let P 1 and P 2 be the two tabulated probabilities with corresponding z values, z 1 and z 2 . Consider 01 21 zz which is the proportion of the distance from z 1 to z 0 . 3 Multiply ( ) PP to obtain a corresponding proportion for the probabilities and add this value to P 1 . This value is the desired ( ) = . Thus, in this case, 1.645 1.64 .005 1 1.65 1.64 .010 2 = = −− and ( ) ( ) [ ] 0 0 1 1 .9495 .9505 .9495 .9495 .0005 .9500 2 P P P  = = + −= + = +   Note: linear interpolation will not be required by midterm 3 or the final exam.
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hw7.solu - STA 13A HW7 Solutions 6.6 a P ( z &lt; 2.33= A (...

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