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1. The masses of largemouth bass, in kilograms, are normally distributed with a mean of 1.1 and a standard deviation of 0.3. Any bass caught with a mass between 0.8kg and 1.7 kg must be released. What percent of the bass does this represent?

Sagot :

Answer:

81.859%

Explanation:

The mean, μ = 1.1

The standard deviation, σ = 0.3

Any bass caught with a mass between 0.8 kg and 1.7 kg must be released

P(X₁ < x < X₂) = P(0.8 < x < 1.7)

For X₁ = 0.8, calculate the z-value

[tex]\begin{gathered} z_1=\frac{X-\mu}{\sigma} \\ \\ z_1=\frac{0.8-1.1}{0.3} \\ \\ z_1=\frac{-0.3}{0.3} \\ \\ z_1=-1 \\ \end{gathered}[/tex]

For X₂ = 1.7, calculate the z-value

[tex]\begin{gathered} z_2=\frac{X_2-\mu}{\sigma} \\ \\ z_2=\frac{1.7-1.1}{0.3} \\ \\ z_2=\frac{0.6}{0.3} \\ \\ z_2=2 \end{gathered}[/tex]

Using the normal distribution table:

P(-1

Percentage of the bass that will be released = 0.81859 x 100%

Percentage of the bass that will be released = 81.859%

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