1. A magazine reported that 10% of American drivers read the newspaper while driving. If 500...

1. A magazine reported that 10% of American drivers read the newspaper while driving. If 500 drivers are selected at random, find the probability that exactly 50 will admit to reading the newspaper while driving.

Expert Solution

Mean ( X ) = n * P = 500 * 0.10 = 50

Standard deviation ( X ) =

Using Normal Approximation to Binomial

P ( X = 50 ) Using continuity correction

P ( 50 - 0.5 < X < 50 + 0.5 ) = P ( 49.5 < X < 50.5 )

P ( 49.5 < X < 50.5 )
Standardizing the value

Z = ( 49.5 - 50 ) / 6.7082
Z = -0.07
Z = ( 50.5 - 50 ) / 6.7082
Z = 0.07
P ( -0.07 < Z < 0.07 )
P ( 49.5 < X < 50.5 ) = P ( Z < 0.07 ) - P ( Z < -0.07 )
P ( 49.5 < X < 50.5 ) = 0.5297 - 0.4703
P ( 49.5 < X < 50.5 ) = 0.0594

P ( X = 50 ) = 0.0594

orchestra answered 3 years ago

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