The diameters of red delicious apples of an orchard have a normal distribution with a mean of 3 inches and a standard deviation of 0.5 inch. What diameter measurement separates the smallest 33% from the largest 67%?

Solution

Given that $\mu = 3$ and $\sigma = 0.5$.

Let the diameter measurement separates the smallest 33% from the largest 67% be $a$

$$ \begin{aligned} & P(X\leq a) =0.33\\ &\Rightarrow P\big(\frac{X-\mu}{\sigma}\leq \frac{a-3}{0.5}\big)=0.33\\ &\Rightarrow P(Z\leq \frac{a-3}{0.5}\big)=0.33\\ &\Rightarrow \frac{a-3}{0.5}= -0.44\\ &\Rightarrow a = 3 + (-0.44)* 0.5\\ &\Rightarrow a = 2.78 \end{aligned} $$

The diameter measurement separates the smallest 33% from the largest 67% is $a = 2.78$.

Further Reading