Question

Suppose that x has a binomial distribution with n = 201 and p = 0.49. (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 (σ) to 4 decimal places.)

(a) Show that the normal approximation to the binomial can appropriately be used to calculate probabilities about x.

np
n(1 – p)

Both np and n(1 – p) ≤ or ≥ 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 = 82)
2.	P (x ≤ 99)
3.	P (x < 74)
4.	P (x ≥ 100)
5.	P (x > 104)

Answer 1

Mean = n * P = ( 201 * 0.49 ) = 98.49
Variance = n * P * Q = ( 201 * 0.49 * 0.51 ) = 50.2299
Standard deviation = = 7.0873

np = 201 * 0.49 = 98.49 > 5

nq = 201 * ( 1 - 0.49 ) = 102.51 > 5

both condition are satisfied.

part 1

P ( X = 82 )
Using continuity correction
P ( n - 0.5 < X < n + 0.5 ) = P ( 82 - 0.5 < X < 82 + 0.5 ) = P ( 81.5 < X < 82.5 )

P ( 81.5 < X < 82.5 )
Standardizing the value

Z = ( 81.5 - 98.49 ) / 7.0873
Z = -2.4
Z = ( 82.5 - 98.49 ) / 7.0873
Z = -2.26
P ( -2.4 < Z < -2.26 )
P ( 81.5 < X < 82.5 ) = P ( Z < -2.26 ) - P ( Z < -2.4 )
P ( 81.5 < X < 82.5 ) = 0.012 - 0.0083
P ( 81.5 < X < 82.5 ) = 0.0038

Part 2

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

P ( X < 99.5 )
Standardizing the value

Z = ( 99.5 - 98.49 ) / 7.0873
Z = 0.14

P ( X < 99.5 ) = P ( Z < 0.14 )
P ( X < 99.5 ) = 0.5557

Part 3

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

P ( X < 73.5 )
Standardizing the value

Z = ( 73.5 - 98.49 ) / 7.0873
Z = -3.53

P ( X < 73.5 ) = P ( Z < -3.53 )
P ( X < 73.5 ) = 0.0002

Part 4

P ( X >= 100 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 100 - 0.5 ) =P ( X > 99.5 )

P ( X > 99.5 ) = 1 - P ( X < 99.5 )
Standardizing the value

Z = ( 99.5 - 98.49 ) / 7.0873
Z = 0.14

P ( Z > 0.14 )
P ( X > 99.5 ) = 1 - P ( Z < 0.14 )
P ( X > 99.5 ) = 1 - 0.5557
P ( X > 99.5 ) = 0.4443

Part 5

P ( X > 104 )
Using continuity correction
P ( X > n + 0.5 ) = P ( X > 104 + 0.5 ) = P ( X > 104.5 )

P ( X > 104.5 ) = 1 - P ( X < 104.5 )
Standardizing the value

Z = ( 104.5 - 98.49 ) / 7.0873
Z = 0.85

P ( Z > 0.85 )
P ( X > 104.5 ) = 1 - P ( Z < 0.85 )
P ( X > 104.5 ) = 1 - 0.8023
P ( X > 104.5 ) = 0.1977