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MA 132 Applications of the Normal Distribution Presentation

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MA 132 Section 7.2 Applications of the Normal Distribution !-Scores n "-scores represent the distance that a data value is from the mean in terms of the number of standard deviations. $−& "= ' Calculating a z-score n n n n n n IQ scores are normally distributed with a mean ! = 100 and % = 15 . Calculate the z-score for an IQ of 120 '= ()* + = ,-.),.. ,/ = 1.33 What does this mean? An IQ of 120 is 1.33 “standard deviations” above the mean. i.e., 120 ≈ 100 + 1.33×15 Finding Probabilities/Proportions Using a Normal Curve r u PIGS ) EXE 80 • 8536 2nd VARS, - ! ! ! % 2 : normal Cdf normakdf ( 65 , 80 , 70,45) Henniker .. . ÷ : : Example, continued - 1333 PIX 275) / 70 normal Cdf ( 75, = y 75 to IE 99 . I 1333 E 99 70 , , ) 4. S Example, continued . 0377 Norma rcdfl = I - - D I E.99 62 70 . - IE 99 0377 , 62 , 70 , 4. s ) Example, continued 75.77 2nd 3 vars . inv Norm inv Norm Area is • go f. 9,0 , u Area . to the \ left X H 75.77 of X M T 70 4. s , ) Example 2: Suppose we know that the heights of men aged 18-24 follow a normal distribution with mean ! = 70.0 inches and standard deviation & = 2.8 inches n What proportion of all adult men aged 18-24 is less than 73 inches tall? 2nd vars - IE99 g / 70 73 M 2. ) normalcdff-IF.gg?,?pep7n92g8 = . 8580 Example 2: Suppose we know that the heights of men aged 18-24 follow a normal distribution with mean ! ...
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