Question

1. How many total students were surveyed 2. Which number of pets was surveyed the leasest 3. P(student with 2 pets)4. P(student in 8th grade)5. P(student in 7th grade or has 1 pet)6. P(student in 6th grade with 3 pets)7. P(student has 1 pet and in 7th grade)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

1) 150 students were surveyed.

2) From the table sketched above we can deduce that 4 pets were surveyed the leasest.

`\text{Probability}=\frac{Number\text{ of required outcomes}}{Total\text{ number of outcomes}}`

3) The total number of students with 2pets= 39

while the total number of pets of all the grades= 150

`P(\text{student with 2pets)=}(39)/(150)=\text{ }\frac{\text{13}}{50}=0.26`

4) P(student with 8th grade)= ?

The total number of student in 8th grade= 50

while the total number of students= 150

`\begin{gathered} P(\text{student in the 8th grade)= }(50)/(150)=(1)/(3)\text{ } \\ =\text{ 0.33} \end{gathered}`

5) The total number of students in 7th grade=50

Total number of students= 150

`\begin{gathered} P(\text{student in 7th grade or has 1 pet)= P(student in 7th grade)} \\ \text{plus the P( student with 1 pet)} \end{gathered}`

The total number of student that has one pet= 62

`\begin{gathered} P(7th\text{ grade)= }(50)/(150)=(1)/(3)=\text{ 0.33} \\ P(1\text{pet)}=\text{ }\frac{\text{62}}{150}=\text{ 0.41} \\ P(7th\text{ grade and 1pet) = }(22)/(150) \\ P(7th\text{ grade or 1 pet)= 0.33+0.41 - 0.15= }0.59 \end{gathered}`

6) Total number of students in the 6th grade with 3pets is a conditional statement.

`\begin{gathered} P(6th\text{ grade with 3pets)=}\frac{P(6th\text{ grade}\cap3pets)}{\text{Total number of 3pets}} \\ =\text{ }(5)/(10)=(1)/(2)=\text{ 0.5} \end{gathered}`

7) Total number of students that has 1 pet and in 7th grade is a conditional statement.

`\begin{gathered} P(1\text{ pet and in 7th grade)}=\frac{P(1pet\cap7thgrade)}{\text{Total number of students in the 7th grade}} \\ =\text{ }(22)/(50)=0.44 \end{gathered}`