Z-score test

The fish weights in a certain lake are normally distributed with a mean of 11 lb and a standard deviation of 6. If four fish are randomly selected, what is the probability that the mean weight will be between 8.6 and 14.6 lb? Your answer should be a decimal rounded to the fourth decimal place.

Correct answer:

p =  0.673

Step-by-step explanation:

$$\begin{gathered} \mu =11\text{lb} \\ \sigma =6\text{lb} \\ n=4 \\ s=\sigma /\sqrt{n}=6/\sqrt{4}=3\text{lb} \\ z=\frac{X-\mu }{s} \\ {x}_1=8.6\text{lb} \\ {x}_2=14.6\text{lb} \\ {z}_1=\frac{x_1-\mu }{s}=\frac{8.6-11}{3}=-\frac{4}{5}=-0.8 \\ {z}_2=\frac{x_2-\mu }{s}=\frac{14.6-11}{3}=\frac{6}{5}=1\frac{1}{5}=1.2 \\ {p}_1=N(z<z_1) \\ {p}_1=0.2119 \\ {p}_2=N(z<z_2) \\ {p}_2=0.8849 \\ p=p_2-p_1=0.8849-0.2119=\frac{673}{1000}=0.673 \end{gathered}$$

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