Question

The labor force participation rate is approximately 63% in the US (i.e. approximately

63% of eligible adults actually work). Suppose 100 adults are randomly selected (assume

independence).

(a) Use the binomial distribution to find the probability that exactly 70 work.

(b) Use the normal approximation to find the probability that exactly 70 work.

(c) Use the normal approximation to find the probability that 70 or less work.

Answer 1

Part a)

X ~ B ( n = 100 , P = 0.63 )

Part b)

Using Normal Approximation to Binomial
Mean = n * P = ( 100 * 0.63 ) = 63
Variance = n * P * Q = ( 100 * 0.63 * 0.37 ) = 23.31
Standard deviation = √(variance) = √(23.31) = 4.828

P ( X = 70 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 70 - 0.5 < X < 70 + 0.5 ) = P ( 69.5 < X < 70.5 )

X ~ N ( µ = 63 , σ = 4.828 )
P ( 69.5 < X < 70.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 69.5 - 63 ) / 4.828
Z = 1.35
Z = ( 70.5 - 63 ) / 4.828
Z = 1.55
P ( 1.35 < Z < 1.55 )
P ( 69.5 < X < 70.5 ) = P ( Z < 1.55 ) - P ( Z < 1.35 )
P ( 69.5 < X < 70.5 ) = 0.9394 - 0.9115
P ( 69.5 < X < 70.5 ) = 0.0279

Part c)

P ( X <= 70 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 70 + 0.5 ) = P ( X < 70.5 )

X ~ N ( µ = 63 , σ = 4.828 )
P ( X < 70.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 70.5 - 63 ) / 4.828
Z = 1.55
P ( ( X - µ ) / σ ) < ( 70.5 - 63 ) / 4.828 )
P ( X < 70.5 ) = P ( Z < 1.55 )
P ( X < 70.5 ) = 0.9394