The average length of the fish caught in lake amotan is 12.3 in. with a standard deviation of 4.1 in. assuming normal distribution, find the probability that a fish caught there will be longer than 18in.

the probability that a fish caught there will be longer than 18in is 0.08226

Step-by-step explanation:

Assuming a normal distribution for length of the fishes caught in Lake Amotan, the formula for normal distribution is expressed as

z = (x - µ)/σ

Where

x = length of fishes

µ = mean length

σ = standard deviation

From the information given,

µ = 12.3 in

σ = 4.1 in

We want to find the probability that a fish caught there will be longer than 18in. It is expressed as

P(x > 18) = 1 - P(x ≤ 18)

For x = 18,

z = (18 - 12.3)/4.1 = 1.39

Looking at the normal distribution table, the probability corresponding to the z score is 0.91774

P(x > 18) = 1 - 0.91774 = 0.08226

(1,4) is on graph of f(x)

f(1) = 4

so 1/4 f(x)

1/4 f(1)

1/4. ×4

1

so point for function 1/4. f(x) is (1,1)

step-by-step explanation: