Respuesta :
Answer:
0.082 is the probability that a fish caught will be longer than 18 inch.
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 12.3 inch
Standard Deviation, σ = 4.1 inch
We are given that the distribution of average length of the fish is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
a) P( longer than 18 inch)
P(x > 18)
[tex]P( x > 18) = P( z > \displaystyle\frac{18 - 12.3}{4.1}) = P(z > 1.39)[/tex]
[tex]= 1 - P(z \leq 1.39)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 18) = 1-0.918 =0.082[/tex]
0.082 is the probability that a fish caught will be longer than 18 inch.
Answer: 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
