Question

In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal distribution to estimate the requested probabilities.

Ocean fishing for billfish is very popular in the Cozumel region of Mexico. In the Cozumel region about 41% of strikes (while trolling) resulted in a catch. Suppose that on a given day a fleet of fishing boats got a total of 21 strikes. Find the following probabilities. (Round your answers to four decimal places.)

(a) 12 or fewer fish were caught
  

(b) 5 or more fish were caught


(c) between 5 and 12 fish were caught

Answer 1

To check for the Normal approximation condition

np > 5 and nq > 5

np = 21 * 0.41 = 8.61 > 5

nq = 21 * ( 1 - 0.41 ) = 12.39 > 5

hence condition is satisfied, we can use Normal approximation to Binomial

Mean = n * P = ( 21 * 0.41 ) = 8.61
Variance = n * P * Q = ( 21 * 0.41 * 0.59 ) = 5.0799
Standard deviation =   = 2.2539

Part a)

P ( X <= 12 )
Using continuity correction
P ( X < n + 0.5 ) = P ( X < 12 + 0.5 ) = P ( X < 12.5 )

P ( X < 12.5 )
Standardizing the value

Z = ( 12.5 - 8.61 ) / 2.2539
Z = 1.73

P ( X < 12.5 ) = P ( Z < 1.73 )
P ( X < 12.5 ) = 0.9582

Part b)

P ( X >= 5 )
Using continuity correction
P ( X > n - 0.5 ) = P ( X > 5 - 0.5 ) =P ( X > 4.5 )

P ( X > 4.5 ) = 1 - P ( X < 4.5 )
Standardizing the value

Z = ( 4.5 - 8.61 ) / 2.2539
Z = -1.82

P ( Z > -1.82 )
P ( X > 4.5 ) = 1 - P ( Z < -1.82 )
P ( X > 4.5 ) = 1 - 0.0344
P ( X > 4.5 ) = 0.9656

part c)

P ( 5 < X < 12 )
Using continuity correction
P ( n + 0.5 < X < n - 0.5 ) = P ( 5 + 0.5 < X < 12 - 0.5 ) = P ( 5.5 < X < 11.5 )

P ( 5.5 < X < 11.5 )
Standardizing the value

Z = ( 5.5 - 8.61 ) / 2.2539
Z = -1.38
Z = ( 11.5 - 8.61 ) / 2.2539
Z = 1.28
P ( -1.38 < Z < 1.28 )
P ( 5.5 < X < 11.5 ) = P ( Z < 1.28 ) - P ( Z < -1.38 )
P ( 5.5 < X < 11.5 ) = 0.9001 - 0.0838
P ( 5.5 < X < 11.5 ) = 0.8163