Problem Statement
The weights of bags of rice are normally distributed. $10\%$ of bags weigh less than $4.8\text{ kg}$ and $5\%$ weigh more than $5.5\text{ kg}$. Find the mean $\mu$ and standard deviation $\sigma$.
Verified Solution & Marking Scheme
Inverse Normal Z-Scores
$P(Z < z_1) = 0.10 \implies z_1 = -1.2816$.
$P(Z < z_2) = 0.95 \implies z_2 = 1.6449$.
Simultaneous Equations
$4.8 - \mu = -1.2816\sigma \quad \text{and} \quad 5.5 - \mu = 1.6449\sigma$
Solve for σ and μ
$5.5 - 4.8 = (1.6449 - (-1.2816))\sigma \implies 0.7 = 2.9265\sigma \implies \sigma \approx 0.239\text{ kg}$
$\mu = 5.5 - 1.6449(0.239) \approx 5.11\text{ kg}$