Question

Assume that only 75% of all drivers in a state regularly wear a seat belt. If a random sample of 400 conductors is selected, what is the probability that: a) There are between 275 and 325 drivers who wear a seat belt regularly b) 300 or fewer sample drivers wear a seat belt regularly. c) At least 290 drivers wear their belts regularly.

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 400 * 0.75 ) = 300
Variance = n * P * Q = ( 400 * 0.75 * 0.25 ) = 75
Standard deviation = √(variance) = √(75) = 8.6603

Part a)
P ( 275 <= X <= 325 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 275 - 0.5 < X < 325 + 0.5 ) = P ( 274.5 < X < 325.5 )

X ~ N ( µ = 300 , σ = 8.6603 )
P ( 274.5 < X < 325.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 274.5 - 300 ) / 8.6603
Z = -2.94
Z = ( 325.5 - 300 ) / 8.6603
Z = 2.94
P ( -2.94 < Z < 2.94 )
P ( 274.5 < X < 325.5 ) = P ( Z < 2.94 ) - P ( Z < -2.94 )
P ( 274.5 < X < 325.5 ) = 0.9984 - 0.0016
P ( 274.5 < X < 325.5 ) = 0.9967

Part b)
P ( X <= 300 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 300 + 0.5 ) = P ( X < 300.5 )

X ~ N ( µ = 300 , σ = 8.6603 )
P ( X < 300.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 300.5 - 300 ) / 8.6603
Z = 0.06
P ( ( X - µ ) / σ ) < ( 300.5 - 300 ) / 8.6603 )
P ( X < 300.5 ) = P ( Z < 0.06 )
P ( X < 300.5 ) = 0.5239

Part c)
P ( X >= 290 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 290 - 0.5 ) =P ( X > 289.5 )

X ~ N ( µ = 300 , σ = 8.6603 )
P ( X > 289.5 ) = 1 - P ( X < 289.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 289.5 - 300 ) / 8.6603
Z = -1.21
P ( ( X - µ ) / σ ) > ( 289.5 - 300 ) / 8.6603 )
P ( Z > -1.21 )
P ( X > 289.5 ) = 1 - P ( Z < -1.21 )
P ( X > 289.5 ) = 1 - 0.1131
P ( X > 289.5 ) = 0.8869