Suppose that x has a binomial distribution with n
= 202 and p = 0.47. (Round np and n(1p) 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)  

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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