Question

The weight W of fish in a given pond is normally distributed with mean 8.5 pounds and standard deviation 1.2 pounds.

(a) What is the probability that a fish weighs less than 8 pounds?

(b) The weight of 90% of fish is below what value?

(c) If you randomly select 5 fish from the pond, what is the probability that the mean weight of the fish is between 8 and 9 pounds?

Answer 1

Part a)

X ~ N ( µ = 8.5 , σ = 1.2 )
P ( X < 8 )
Standardizing the value
Z = ( X - µ ) / σ
Z = ( 8 - 8.5 ) / 1.2
Z = -0.4167
P ( ( X - µ ) / σ ) < ( 8 - 8.5 ) / 1.2 )
P ( X < 8 ) = P ( Z < -0.4167 )
P ( X < 8 ) = 0.3384

Part b)

X ~ N ( µ = 8.5 , σ = 1.2 )
P ( X < x ) = 90% = 0.9
To find the value of x
Looking for the probability 0.9 in standard normal table to calculate Z score = 1.2816
Z = ( X - µ ) / σ
1.2816 = ( X - 8.5 ) / 1.2
X = 10.0379 ≈ 10.0
P ( X < 10.0379 ) = 0.9

Part c)

X ~ N ( µ = 8.5 , σ = 1.2 )
P ( 8 < X < 9 )
Standardizing the value
Z = ( X - µ ) / ( σ / √(n))
Z = ( 8 - 8.5 ) / ( 1.2 / √(5))
Z = -0.9317
Z = ( 9 - 8.5 ) / ( 1.2 / √(5))
Z = 0.9317
P ( -0.93 < Z < 0.93 )
P ( 8 < X̅ < 9 ) = P ( Z < 0.93 ) - P ( Z < -0.93 )
P ( 8 < X̅ < 9 ) = 0.8243 - 0.1757
P ( 8 < X̅ < 9 ) = 0.6485