Express probabilities in terms of z

\( P(X < 490) = 0.05 \Rightarrow z_1 = \text{invNorm}(0.05) \approx -1.645 \)
\( P(X > 510) = 0.10 \Rightarrow P(X < 510) = 0.90 \Rightarrow z_2 = \text{invNorm}(0.90) \approx 1.281 \)

Set up equations

\( \frac{490 – \mu}{\sigma} = -1.645 \quad \Rightarrow 490 – \mu = -1.645 \sigma \)
\( \frac{510 – \mu}{\sigma} = 1.281 \quad \Rightarrow 510 – \mu = 1.281 \sigma \)

Solve the system

Subtract the two equations:
\( (510 – \mu) – (490 – \mu) = 1.281 \sigma + 1.645 \sigma \)
\( 20 = 2.926 \sigma \Rightarrow \sigma = 20 / 2.926 \approx 6.84 \)

Use \( \sigma \) to find \( \mu \):
\( 490 – \mu = -1.645 \times 6.84 \Rightarrow 490 – \mu = -11.25 \Rightarrow \mu = 501.25 \)

Final answer:

Mean \( \mu \approx 501.25 \) g, Standard deviation \( \sigma \approx 6.84 \) g

Translate probabilities into z-values

For \( P(X < 60) = 0.20 \):
Use GDC: invNorm(0.20) → \( z_1 \approx -0.8416 \)

For \( P(X > 85) = 0.10 \Rightarrow P(X < 85) = 0.90 \):
Use GDC: invNorm(0.90) → \( z_2 \approx 1.2816 \)

Set up equations using z formula

\( \frac{60 – \mu}{\sigma} = -0.8416 \Rightarrow 60 – \mu = -0.8416 \sigma \)
\( \frac{85 – \mu}{\sigma} = 1.2816 \Rightarrow 85 – \mu = 1.2816 \sigma \)

Subtract:
\( 85 – 60 = 1.2816 \sigma + 0.8416 \sigma \)
\( 25 = 2.1232 \sigma \Rightarrow \sigma = 25 / 2.1232 \approx 11.78 \)

\( 60 – \mu = -0.8416 \times 11.78 \Rightarrow 60 – \mu = -9.91 \Rightarrow \mu = 60 + 9.91 = 69.91 \)

Final result:

Mean \( \mu \approx 69.9 \), standard deviation \( \sigma \approx 11.8 \)