The random variable X~(30,2^2)
Find p(X<33)
Find p(X>26)

Answer:

i) P(X<33)  = 0.9232

ii) P(X>26) = 0.001

Step-by-step explanation:

Step(i):-

Given that the mean of the Population = 30

Given that the standard deviation of the Population = 4

Let 'X' be the Normal distribution

Step(ii):-

i)

Given that the random variable  X = 33

               [tex]Z = \frac{x-mean}{S.D}[/tex]

               [tex]Z = \frac{33-30}{2} = 1.5[/tex] >0

P(X<33) = P( Z<1.5)

              = 1- P(Z>1.5)

             = 1 - ( 0.5 - A(1.5))

             = 0.5 + 0.4232

  P(X<33)  = 0.9232

Step(iii) :-

Given that the random variable  X = 26

               [tex]Z = \frac{x-mean}{S.D}[/tex]

               [tex]Z = \frac{33-26}{2} = 3.5[/tex] >0

P(X>26)  = P( Z>3.5)

              = 0.5 - A(3.5)

              = 0.5 - 0.4990

             = 0.001

P(X>26) = 0.001

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