Assume that the random variable $X$ $X$ is normally distributed, with mean = 50 and standard deviation = 7. How would I compute the probability $P (X > 35)$ $P( X > 35 )$ ?

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Answer 1 · 2017-05-09T16:22:20

$P (Z > - 2.14) = 0.9838$ $P(Z> -2.14)=0.9838$

$X ~ N (50, 7^{2})$ $X~N(50,7^2)$

$P (X > 35)$ $P(X>35)$

standardising

$P (X > 35) \to P (Z > \frac{35 - 50}{7})$ $P(X>35)rarrP(Z>(35-50)/7)$

$= P (Z > - 2.14) (2 d p s)$ $=P(Z> -2.14)" ( 2dps)$

now by the symmetry of the Normal curve

$P (Z > - 2.14) = P (Z < 2.14)$ $P(Z> -2.14)=P(Z<2.14)$

so we now look up directly from tables

$P (Z > - 2.14) = 0.9838$ $P(Z> -2.14)=0.9838$

Assume that the random variable XXX is normally distributed, with mean = 50 and standard deviation = 7. How would I compute the probability P( X > 35 )P(X>35)P( X > 35 )?