Question

The weights of the fish in a certain lake are normally distributed with a mean of 12 lb and a standard deviation of 12. If 16 fish are randomly selected, what is the probability that the mean weight will be between 9.6 and 15.6lb?

Answer 1

Answer

The probability that the mean weight will be between 9.6 and 15.6lb = 0.1972

Explanation

Mean, μ = 12 lb

Standard deviation, δ = 12

Number of population, n = 16

Let x₁ = 9.6 lb and x₂ = 15 .6 lb

The z score formula is given by;

$z=\frac{x-\mu}{\sigma}$$\begin{gathered} z_{_1}=\frac{x_1-\mu}{\sigma}=\frac{9.6-12}{12} \\ z_{_1}=-\frac{2.4}{12} \\ z_{_1}=-0.2 \end{gathered}$

For z₂;

$\begin{gathered} z_2=\frac{X_2-\mu}{\sigma}=\frac{15.6-12}{12} \\ z_2=\frac{3.6}{12} \\ z_2=0.3 \end{gathered}$

We want to find this probability: P(9.6 < X < 15.6)

We can calculate this probability like this:

[tex]\begin{gathered} P(z_1