help!!! .com/al/servlet/quizictk-sbarrett-00338iquiz action-t akeQuiz&quiz; prob

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.com/al/servlet/quizictk-sbarrett-00338iquiz action-t akeQuiz&quiz; probGuid-QNAPCOA8010100 Body Tail Body Tail Mean to ž Body Tail Mean to z 1.500.93320.0668 0.4332 2100.9821 2.75 2.85 0.9978 0.0022 0.20 0.5793 0.42070.0793 0.55 0.0179 0.4821 1.15 0.8749 0.125 0.3749 1.35 0.9115 0.088504115 0.4978 To find the probablity of a z-score, position the orange line at the appropriate z-score on the horizontal axis. The areas under the standard normal curve to the left and right of the vertical line are displayed in blue and orange, respectively (Hint: The standard normal distribution is symmetrical about the mean, the area under the curve to the ieft (and right) of the mean is 0.5. Therefore, the area that corresponds with Mean to z is computed as Larger Portion (Body or Tain) 0.S.) Standard Norcial Distrbution Standard Deviationnt 5000 002 SC2501.n o s0.oo .s0 1.001.50 2.00 2.50 3.00

Explanation / Answer

For z = 0.55

P( Body) , Use excel function

NORMSDIST(0.55)

= 0.7088

P( Tail)

= 1-0.7088

= 0.2912

Mean to z = 0.7088-0.50 = 0.2088

z Body Tail Mean to z  

0.55 0.7088 0.2912 0.2088

For z = 2.75

P( Body) , Use excel function

NORMSDIST(2.75)

= 0.9970

P( Tail)

= 1-0.9970

= 0.0030

Mean to z = 0.9970-0.50 = 0.4970

z Body Tail Mean to z  

0.55 0.9970 0.0030 0.4970

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