Question

There were 49.7 million people with some type of long-lasting condition or disability living in the United States in 2000. This represented 19.3 percent of the majority of civilians aged five and over (http://factfinder.census.gov). A sample of 1000 persons is selected at random. Use normal approximation. Round the answers to four decimal places (e.g. 98.7654).

(a) Approximate the probability that more than 201 persons in the sample have a disability. (answer is not .274) round 4 decimal plzzzzz on a and b

(b) Approximate the probability that between 180 and 300 people in the sample have a disability. (answer is not .860

Answer 1

Using Normal Approximation to Binomial
Mean = n * P = ( 1000 * 0.193 ) = 193
Variance = n * P * Q = ( 1000 * 0.193 * 0.807 ) = 155.751
Standard deviation = √(variance) = √(155.751) = 12.48

Part a

P ( X > 201 )
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 201 + 0.5 ) = P ( X > 201.5 )

X ~ N ( µ = 193 , σ = 12.48 )
P ( X > 201.5 ) = 1 - P ( X < 201.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 201.5 - 193 ) / 12.48
Z = 0.68
P ( ( X - µ ) / σ ) > ( 201.5 - 193 ) / 12.48 )
P ( Z > 0.68 )
P ( X > 201.5 ) = 1 - P ( Z < 0.68 )
P ( X > 201.5 ) = 1 - 0.7517
P ( X > 201.5 ) = 0.2483

Part b)

P ( 180 <= X <= 300 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 180 - 0.5 < X < 300 + 0.5 ) = P ( 179.5 < X < 300.5 )

X ~ N ( µ = 193 , σ = 12.48 )
P ( 179.5 < X < 300.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 179.5 - 193 ) / 12.48
Z = -1.08
Z = ( 300.5 - 193 ) / 12.48
Z = 8.61
P ( -1.08 < Z < 8.61 )
P ( 179.5 < X < 300.5 ) = P ( Z < 8.61 ) - P ( Z < -1.08 )
P ( 179.5 < X < 300.5 ) = 1 - 0.1401
P ( 179.5 < X < 300.5 ) = 0.8599