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it is known that X~N(26.9;5.5). calculate p(25<X<38)..

could anyone help this problem?

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Language Literature and Social Science Researcher with 05 years experience

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0.6113
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Btech,JEEQualified,Studied from ALLEN, Kota, deliver lectures on MATHS & SCIENCE for class 10th

0.6113
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Z=P(25-26.9/5.5<Z<38-26.9/5.5) = P(-0.34<Z<2.02) = 0.1331+0.4782 = 0.6113
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Z=P(25-26.9/5.5<Z<38-26.9/5.5)=P(-0.34<Z<2.02)=0.1331+0.4782 =0.6113
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Z=(Mean -X)/SD NOW calculate value of Z and Find the value from normal distribution curve
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Z=X-Mean/standard deviation P(25-26.9/5.5<Z<38-26.9/5.5) =P(-0.34<Z<2.02) =0.1331+0.4782=0.6113 Nice
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Z=P(25-26.9/5.5<Z<38-26.9/5.5)=P(-0.34<Z<2.02)=0.1331+0.4782 =0.6113
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I am a professional trainer od song and class 1 to 12 student.. I have 10 year experience.

Z=P(25-26.9/5.5<Z<38-26.9/5.5)=P(-0.34<Z<2.02)=0.1331+0.4782 =0.6113
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Z=P(25-26.9/5.5<Z<38-26.9/5.5) = P(-0.34<Z<2.02) = 0.1331+0.4782 = 0.6113
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