Question

1. How many total engineers liked science 2. Which type of engineer was surveyed the most 3. P(mechanical engineer)4. P(engineer who liked math)5. P(engineer who liked science or is a chemical engineer)6. P(electrical or chemical engineer)7. P(mechanical engineer who liked science)8. P(engineer who liked math and is an electrical engineer)Answer the following problems about two way frequency tables fill in the missing cells of each table make sure to reduce your fraction.

Answer 1

Step 1; Complete the table

1. The number of engineers that like science are 266

2. The engineer surveyed the most is Electrical Engineers

3. Pr(mechanical Engineer) is

`\begin{gathered} \frac{total\text{ number of mechanical engin}er}{\text{total}} \\ \text{Total number pf mechanical engine}er\text{ =170} \\ \text{Total =516} \\ =(170)/(516) \\ =0.33 \end{gathered}`

4. Pr(engineer who liked maths ) is

`\frac{Total\text{ number of engin}er\text{ who liked maths }}{ground\text{ total}}=(254)/(516)=0.49`

5. Pr( Engineer who like science or is a chemical engineer) is

Let Pr(engineer like science)=E and Pr(a chemical engineer)=S

The Pr( Engineer who like science or is a chemical engineer) is given as

`\begin{gathered} Pr(E)+Pr(S)-Pr(\text{EnS)} \\ \text{where Pr(E)=}(262)/(516),\text{ Pr(S)=}(171)/(516),\text{ Pr(EnS)=}(91)/(516) \end{gathered}`

Hence, we have

`\begin{gathered} (262)/(516)+(171)/(516)-(91)/(516) \\ =(262+171-91)/(516)=(342)/(516)=0.66 \end{gathered}`

6. Pr( Electrical or chemical) is

`\begin{gathered} Pr(\text{Electrical) +Pr(chemical)} \\ Pr(\text{electrical)}=(175)/(516),\text{ Pr(chemical)}=(171)/(516) \\ \text{hence,} \\ (175)/(516)+(171)/(516)=(346)/(516)=0.67 \\ \end{gathered}`

7. Pr(Mechanical engineer who liked science) is given as

`\begin{gathered} Pr(M|S)=(Pr(MnS))/(Pr(S)) \\ Pr(MnS)=(81)/(516),\text{ Pr(S)=}(262)/(516) \\ Pr(M|S)=(81)/(516)*(516)/(262)=(81)/(262)=0.31 \end{gathered}`

8. Pr(an engineer who liked maths and is an electrical engineer)

`(85)/(175)=0.49`