A manufacturer produces 25-pound lifting weights. The lowest actual weight is 24 pounds, and the highest is 26 pounds. Each weight is equally likely so the distribution of weights is uniform. A sample of 100 weights is taken. Find the 90th percentile for the total weight of the 100 weights.
Solution
The probability density curve is
$$ \begin{aligned} f(x)=\left\{ \begin{array}{ll} \frac{1}{26 - 24} , & \hbox{$24 \leq x\leq 60$;} \\ 0, & \hbox{Otherwise.} \end{array} \right. \end{aligned} $$
The mean of $X$ is $\mu_x= \frac{24+26}{2}=25$ and variance is $\sigma^2_x = \frac{(26-24)^2}{12}=\frac{4}{12}= 0.3333$. That is the standard deviation is $\sigma_x = 0.5774$.
The distribution of the total actual weight for the 100 weights $(Y=\sum X_i)$ is Normal with mean $\mu =25\times 100 = 2500$ and standard deviation $\sigma = 5.7735$. (using Central Limit Theorem).
Let the $90^{th}$ percentile for the total weight of the 100 weights is $A$.
$$ \begin{aligned} P(Y\leq A) &= 0.90\\ \Rightarrow & P\bigg(\frac{Y-\mu}{\sigma}\leq \frac{A-2500}{5.7735}\bigg) = 0.90\\ \Rightarrow & P\bigg(Z\leq z_1\bigg) = 0.90 \end{aligned} $$
$$ \begin{aligned} z_1 &= \frac{A-2500}{5.7735}= 1.2816\\ \Rightarrow &A =2500 + 5.7735 \times 1.2816\\ \Rightarrow &A =2507.4 \end{aligned} $$
Thus the $90^{th}$ percentile for the total weight of the 100 weights is $2507.4$.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
