Problem Statement
The volumes of soft drink in bottles filled by a factory follow a normal distribution with mean $\mu$ and standard deviation $\sigma$.
It is known that $8\%$ of bottles contain less than $490\text{ mL}$, and $15\%$ of bottles contain more than $515\text{ mL}$.
Find the value of $\mu$ and the value of $\sigma$.
Verified Solution & Marking Scheme
Standardize lower threshold
$$P(X < 490) = 0.08 \implies z_1 = \text{invNorm}(0.08) = -1.40507$$\n$$\frac{490 - \mu}{\sigma} = -1.40507 \implies 490 - \mu = -1.40507\sigma$$ (1)
Standardize upper threshold
$$P(X > 515) = 0.15 \implies P(X < 515) = 0.85 \implies z_2 = \text{invNorm}(0.85) = 1.03643$$\n$$\frac{515 - \mu}{\sigma} = 1.03643 \implies 515 - \mu = 1.03643\sigma$$ (2)
Solve system of linear equations
Subtract (1) from (2): $$(515 - 490) = (1.03643 - (-1.40507))\sigma \implies 25 = 2.4415\sigma \implies \sigma = \frac{25}{2.4415} \approx 10.2\text{ mL}$$\n$$\mu = 515 - 1.03643(10.2396) \approx 504.4\text{ mL}$$