Question

use the normal distribution to approximate the following binomial distribution. a die is rolled 84 times. what is the probability that the number 2 occurs more than 12 times?

Answer 1

The required probability is 1 - 0.330 = 0.6697 .

P ( x= 2 ) = 1 /6  [ p , probability of success ]

q = 1 - ( 1 / 6 )

  = 5 / 6

n = 84

Mean = np

 =  84 * ( 1 / 6 )

 = 14

npq = 14 * ( 5 / 6 )

       = 11.667

Applying normal approximation ,

To calculate P ( x < 12 ) = P { Z < [ ( 12 - 14 ) / ( $11.667^{1/2}$ ) ] }

                                       = P [ Z < ( - 2 ) / ( 3.416 ) ]

                     P  [ Z < - 0.585 ] = 0.279

Therefore , the probability 2 occurs more than 12 times = 1 - 0.279

                                                                                      = 0.721

Applying continuity correction ,

P ( x < = 12 ) = P ( x < 12.0 + 0.5 ) = P ( x < 12.5 )

                                                     = P [ Z < ( 12.5 - 14 ) / 3.416 ]

                                               = P ( Z < - 0.439 )

                                                  = 0.330

Hence , the required probability is 1 - 0.330 = 0.6697 .

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