Question

Suppose that x has a binomial distribution with n = 198 and p = 0.44. (Round np and n(1-p) answers to 2 decimal places. Round your answers to 4 decimal places. Round z values to 2 decimal places. Round the intermediate value (o) to 4 decimal places.) (a) Show that the normal approximation to the binomial can appropriately be used to calculate probabilities about x пр n(1 - p) Both np and n(1 – p) (Click to select) A 5 (b) Make continuity corrections for each of the following, and then use the normal approximation to the binomial to find each probability: 1. P (x = 79) 2. P (x s 99) 3. P(x < 74) 4. P(x2 102) 5. P(x > 101)

Answer 1

X ~ B ( n = 198 , P = 0.44 )

Part a)

np = 198 * 0.44 = 87.12

n ( 1 - p ) = 198 * ( 1 - 0.44 ) = 110.88

Both np and nq > 5

Part b)

Using Normal Approximation to Binomial
Mean = n * P = ( 198 * 0.44 ) = 87.12
Variance = n * P * Q = ( 198 * 0.44 * 0.56 ) = 48.7872
Standard deviation = √(variance) = √(48.7872) = 6.9848

1.
P ( X = 79 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 79 - 0.5 < X < 79 + 0.5 ) = P ( 78.5 < X < 79.5 )

X ~ N ( µ = 87.12 , σ = 6.9848 )
P ( 78.5 < X < 79.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 78.5 - 87.12 ) / 6.9848
Z = -1.23
Z = ( 79.5 - 87.12 ) / 6.9848
Z = -1.09
P ( -1.23 < Z < -1.09 )
P ( 78.5 < X < 79.5 ) = P ( Z < -1.09 ) - P ( Z < -1.23 )
P ( 78.5 < X < 79.5 ) = 0.1379 - 0.1093
P ( 78.5 < X < 79.5 ) = 0.0285

2.
P ( X <= 99 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 99 + 0.5 ) = P ( X < 99.5 )

X ~ N ( µ = 87.12 , σ = 6.9848 )
P ( X < 99.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 99.5 - 87.12 ) / 6.9848
Z = 1.77
P ( ( X - µ ) / σ ) < ( 99.5 - 87.12 ) / 6.9848 )
P ( X < 99.5 ) = P ( Z < 1.77 )
P ( X < 99.5 ) = 0.9616

3.
P ( X < 74 )
Using continuity correction
P ( X < n - 0.5 ) = P ( X < 74 - 0.5 ) = P ( X < 73.5 )

X ~ N ( µ = 87.12 , σ = 6.9848 )
P ( X < 73.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 73.5 - 87.12 ) / 6.9848
Z = -1.95
P ( ( X - µ ) / σ ) < ( 73.5 - 87.12 ) / 6.9848 )
P ( X < 73.5 ) = P ( Z < -1.95 )
P ( X < 73.5 ) = 0.0256

4.
P ( X >= 102 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 102 - 0.5 ) =P ( X > 101.5 )

X ~ N ( µ = 87.12 , σ = 6.9848 )
P ( X > 101.5 ) = 1 - P ( X < 101.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 101.5 - 87.12 ) / 6.9848
Z = 2.06
P ( ( X - µ ) / σ ) > ( 101.5 - 87.12 ) / 6.9848 )
P ( Z > 2.06 )
P ( X > 101.5 ) = 1 - P ( Z < 2.06 )
P ( X > 101.5 ) = 1 - 0.9803
P ( X > 101.5 ) = 0.0197

5.
P ( X > 101 )
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 101 + 0.5 ) = P ( X > 101.5 )

X ~ N ( µ = 87.12 , σ = 6.9848 )
P ( X > 101.5 ) = 1 - P ( X < 101.5 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 101.5 - 87.12 ) / 6.9848
Z = 2.06
P ( ( X - µ ) / σ ) > ( 101.5 - 87.12 ) / 6.9848 )
P ( Z > 2.06 )
P ( X > 101.5 ) = 1 - P ( Z < 2.06 )
P ( X > 101.5 ) = 1 - 0.9803
P ( X > 101.5 ) = 0.0197