Math / Data & Statistics

Question4.2 A survey was conducted amorig 130 Grade 11 learnets to establlsh which silack chey prefer to eat while they watch television. The results were summarized in the Venn diagrain below. However, some information is missing 4.2.1 If 29 leamers prefer at least two types of snacks, calculate the value of $x$ and $y$ . 4.2.2 Determine the probability that a learner who does not eat nuts will either have another snack or no snack while watching television. (3)

Studdy Solution

STEP 1

1. The total number of learners surveyed is 130.
2. We are given partial counts of learners' preferences for various combinations of snacks.
3. The values of $x$ and $y$ are unknowns to be determined.
4. The sum of all the disjoint regions in the Venn diagram plus the regions represented by $x$ and $y$ must equal 130.

STEP 2

1. Set up the equation for the total number of learners using the Venn diagram.
2. Use the information provided about learners who prefer at least two types of snacks to form another equation.
3. Solve the system of equations to find the values of $x$ and $y$ .
4. Determine the probability that a learner who does not eat nuts will either have another snack or no snack while watching television.
_STEP_1_: High_Level_Step: 1 Set up the equation for the total number of learners using the Venn diagram.
$25 + 10 + 30 + 4 + 6 + 9 + 5 + x + y = 130$ High_Level_Step_Completed: FALSE
_STEP_2_: High_Level_Step: 1 Simplify the equation to solve for $x + y$ .
$89 + x + y = 130$
$x + y = 41$ High_Level_Step_Completed: TRUE
_STEP_3_: High_Level_Step: 2 Use the information that 29 learners prefer at least two types of snacks. This includes the intersections of two circles and the intersection of all three circles.
$(4 + 6 + 9 + 5) + x = 29$
$24 + x = 29$
$x = 5$ High_Level_Step_Completed: TRUE
_STEP_4_: High_Level_Step: 3 Substitute the value of $x$ into the equation $x + y = 41$ to find $y$ .
$5 + y = 41$
$y = 36$ High_Level_Step_Completed: TRUE
_STEP_5_: High_Level_Step: 4 Determine the probability that a learner who does not eat nuts will either have another snack or no snack while watching television. First, find the number of learners who do not eat nuts.
$130 - (10 + 4 + 9 + 5 + x + y) = 130 - (10 + 4 + 9 + 5 + 5 + 36)$
$130 - 69 = 61$ High_Level_Step_Completed: FALSE
_STEP_6_: High_Level_Step: 4 Find the number of learners who do not eat nuts and will either have another snack or no snack.
The number of learners who do not eat nuts and will have another snack or no snack is simply the total number of learners who do not eat nuts, which is 61.
High_Level_Step_Completed: TRUE
_STEP_7_: High_Level_Step: 4 Calculate the probability.
$P = \frac{61}{130}$ High_Level_Step_Completed: TRUE
Solution: The values of $x$ and $y$ are $5$ and $36$ , respectively. The probability that a learner who does not eat nuts will either have another snack or no snack while watching television is:
$\frac{61}{130}$
High_Level_Step_Completed: TRUE

Was this helpful?

Get Studdy

Scan QR code with your device to get Studdy on iOS or Android

QR code

Studdy solves anything!

Download the app

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Parents Influencer program Contact Policy Terms