Question #147977
(Exa) 1. Find what is the probability of value of X being equals <=110 for the normal distribution defined by the average of 92 and standard deviation of 14. Score?

2. What would be the probability of X being >110 mean is 92 and the standard deviation is 14 (Remember total probability is 1!) Score?
1
Expert's answer
2020-12-03T08:04:19-0500

1)


P(X110)=P(Z1109214)=P(Z97)P(X\leq110)=P(Z\leq\dfrac{110-92}{14})=P(Z\leq\dfrac{9}{7})

P(Z1.285714)0.90072860\approx P(Z\leq1.285714)\approx0.90072860

2)


P(X>110)=1P(X110)P(X>110)=1-P(X\leq110)

=1P(Z1109214)=1P(Z97)=1-P(Z\leq\dfrac{110-92}{14})=1-P(Z\leq\dfrac{9}{7})

1P(Z1.285714)10.90072860\approx 1-P(Z\leq1.285714)\approx1-0.90072860

0.09927140\approx0.09927140


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