Let X normally distributed with mean mu = 132 and standard deviation sigma = 35.

Question

Let X normally distributed with mean mu = 132 and standard deviation sigma = 35. Use Find P(X lessthanorequalto 100). (Round "z" value to 2 decimal places and final answer to 4 al places.) P(X lessthanorequalto 100) Find P(95 lessthanorequalto X lessthanorequalto 110). (Round "z" value to 2 decimal places and final answer to 4 decimal places.) P(95 lessthanorequalto X lessthanorequalto 110) Find x such that P(X lessthanorequalto x) = 0.350. (Round "z" value and final answer to 2 decimal places.) x Find x such that P(X > x) = 0.810. (Round "z" value and final answer to 2 decimal places.) x

Explanation / Answer

a) For x = 100, the z-value z = (100- 132) / 35 = -0.914

Hence P(x < 100) = P(z <-0.914) = [area to the left of -0.914] = 0.1804

b)For x = 95 , z = (95 - 132) / 35 = -1.057 and for x = 110, z = (110 - 132) / 35 =-0.63

Hence P(30 < x < 35) = P(-1.057 < z <-0.63) = [area to the left of z =-0.63] - [area to the left of z= -1.057]

= 0.2643 - 0.1453 = 0.119

c) Given P(X<x) =0.350 , using z table

z score = -0.385 = (x-132)/35

  X = 118.5

d) Given P(X>x) =0.810 ,

implies 1-P(X<x) =0.810

P(X<x) = 0.190

using z table , z score = -0.878

z score = -0.878 = (x-132)/35

  X = 101.27

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