Question

# Suppose that x has a binomial distribution with n = 202 and p = 0.47. (Round...

Suppose that x has a binomial distribution with n = 202 and p = 0.47. (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) (Click to select)≥≤ 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 ≤ 99) 3. P (x < 73) 4. P (x ≥ 101) 5. P (x > 103)

#### Homework Answers

Answer #1

n * P = 202 * 0.47 = 94.94

n * Q = 202 * ( 1 - 0.47 ) = 107.76

Both np and n(1 – p) ≥ 5

Mean ( X ) = n * P = 202 * 0.47 = 94.94

Standard deviation (X) =

Part b)

1. P (x = 79) Using continuity correction

P ( 79 - 0.5 < X < 79 + 0.5 ) = P ( 78.5 < X < 79.5 )

P ( 78.5 < X < 79.5 )

Standardizing the value

Z = ( 78.5 - 94.94 ) / 7.0935

Z = -2.32

Z = ( 79.5 - 94.94 ) / 7.0935

Z = -2.18

P ( -2.32 < Z < -2.18 )

P ( 78.5 < X < 79.5 ) = P ( Z < -2.18 ) - P ( Z < -2.32 )

P ( 78.5 < X < 79.5 ) = 0.0148 - 0.0102

P ( 78.5 < X < 79.5 ) = 0.0045

2. P (x ≤ 99)

Using continuity correction

P ( X < 99 + 0.5 ) = P ( X < 99.5 )

P ( X < 99.5 )

Standardizing the value

Z = ( 99.5 - 94.94 ) / 7.0935

Z = 0.64

P ( X < 99.5 ) = P ( Z < 0.64 )

P ( X < 99.5 ) = 0.7389

3.  P (x < 73)

Using continuity correction

P ( X < 73 - 0.5 ) = P ( X < 72.5 )

P ( X < 72.5 )

Standardizing the value

Z = ( 72.5 - 94.94 ) / 7.0935

Z = -3.16

P ( X < 72.5 ) = P ( Z < -3.16 )

P ( X < 72.5 ) = 0.0008

4. P (x ≥ 101)

Using continuity correction

P ( X > 101 - 0.5 ) = P ( X > 100.5 )

P ( X > 100.5 ) = 1 - P ( X < 100.5 )

Standardizing the value

Z = ( 100.5 - 94.94 ) / 7.0935

Z = 0.78

P ( Z > 0.78 )

P ( X > 100.5 ) = 1 - P ( Z < 0.78 )

P ( X > 100.5 ) = 1 - 0.7823

P ( X > 100.5 ) = 0.2177

5. P (x > 103)

Using continuity correction

P ( X > 103 + 0.5 ) = P ( X > 103.5 )

P ( X > 103.5 ) = 1 - P ( X < 103.5 )

Standardizing the value

Z = ( 103.5 - 94.94 ) / 7.0935

Z = 1.21

P ( Z > 1.21 )

P ( X > 103.5 ) = 1 - P ( Z < 1.21 )

P ( X > 103.5 ) = 1 - 0.8869

P ( X > 103.5 ) = 0.1131

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