use the normal distribution to approximate the following binomial distribution. a die is rolled 84 times. what is the probabilit

Question

2 years ago

6

y that the number 2 occurs more than 12 times?

1 answer:

nadezda [96] · Answer 1 · 2023-08-07T00:48:24+03:00

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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