Answer

Solution

1. 1

   $P(\text{diam}<1.8)=0.08\Rightarrow F\bigl(\tfrac{1.8-\mu}{\sigma}\bigr)=0.08$. From the table, $F(-1.4)=1-F(1.4)=0.08$. Hence

2. 2
   $$\frac{1.8-\mu}{\sigma}=-1.4 \;\Longrightarrow\; \mu-1.8=1.4\sigma.\tag{1}$$
3. 3

   $P(\text{diam}>2.4)=0.055\Rightarrow 1-F\bigl(\tfrac{2.4-\mu}{\sigma}\bigr)=0.055$, i.e.\ $F(\cdot)=0.945=1-0.055$. From the table $F(1.6)=1-F(-1.6)=0.945$. Hence

4. 4
   $$\frac{2.4-\mu}{\sigma}=1.6 \;\Longrightarrow\; 2.4-\mu=1.6\sigma.\tag{2}$$
5. 5

   Add (1) and (2): $(2.4-\mu)+(\mu-1.8)=1.6\sigma+1.4\sigma\Rightarrow 0.6=3\sigma\Rightarrow \sigma=0.2$. Then $\mu=1.8+1.4(0.2)=1.8+0.28=2.08$.

Answer structure / marking notes

Content note