Question

1. On average fisherman catch 10,000 fishes a month with standard deviation of 1,500. If fishermen around the world was asked to present their weekly averages. Find the probability that (assume variable is normally distributed):

• Central Limit theorem

1. 100 fishermen catch more than 10,500 fish a week
2. 100 fishermen catch less than 10,100 fish a week
3. 100 fishermen catch between 9,000 and 11,000 fish a week

Answer 1

a)

X ~ N ( µ = 10000 , σ = 1500 )
P ( X > 10500 ) = 1 - P ( X < 10500 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 10500 - 10000 ) / ( 1500 / √ ( 100 ) )
Z = 3.33
P ( ( X - µ ) / ( σ / √ (n)) > ( 10500 - 10000 ) / ( 1500 / √(100) )
P ( Z > 3.33 )
P ( X̅ > 10500 ) = 1 - P ( Z < 3.33 )
P ( X̅ > 10500 ) = 1 - 0.9996 (From Z table)  
P ( X̅ > 10500 ) = 0.0004

b)

X ~ N ( µ = 10000 , σ = 1500 )
P ( X < 10100 )
Standardizing the value
Z = ( X - µ ) / (σ/√(n)
Z = ( 10100 - 10000 ) / ( 1500 / √100 )
Z = 0.67
P ( ( X - µ ) / ( σ/√(n)) = ( 10100 - 10000 ) / ( 1500 / √(100) )
P ( X < 10100 ) = P ( Z < 0.67 )
P ( X̅ < 10100 ) = 0.7486 (From Z table)

c)

X ~ N ( µ = 10000 , σ = 1500 )
P ( 9000 < X < 11000 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 9000 - 10000 ) / ( 1500 / √(100))
Z = -6.67
Z = ( 11000 - 10000 ) / ( 1500 / √(100))
Z = 6.67
P ( 9000 < X̅ < 11000 ) = P ( Z < 6.67 ) - P ( Z < -6.67 )
P ( 9000 < X̅ < 11000 ) = 1 - 0 (From Z table)
P ( 9000 < X̅ < 11000 ) = 1