Weighted Average - Examples, Exercises and Solutions

In order to determine the hotel rating, we must calculate an average.

Below is the weighted average formula:

(value A X weight percentage A)+(value B X weight percentage B)...

First, let's add up all the percentages:

$50\%+30\%+10\%+10\%=100\%$

Now we must multiply each factor by its weight percentage, convert the percentages to decimal numbers, and add them as follows:

$4.5\times0.5+4\times0.3+5\times0.1+3\times0.1=$

We then proceed to solve the multiplication exercises:

$2.25+1.2+0.5+0.3=$

Finally we add them up and obtain the following: 4.25 which is the hotel's overall rating.

Let's solve the problem by applying the steps we outlined:

• Step 1: Identify the grade and weight for each component:
    - Assignment: Grade = 98, Weight = 30% (or 0.30)
    - Exam: Grade = 95, Weight = 70% (or 0.70)
• Step 2: Apply the weighted average formula:
    - Compute the weighted grade for each component:
      Weighted grade for assignment = $98 \times 0.30 = 29.4$
      Weighted grade for exam = $95 \times 0.70 = 66.5$
• Step 3: Sum the weighted grades to find Monica's average grade:
    - Total weighted average = $29.4 + 66.5 = 95.9$

Therefore, Monica's average grade is $95.9$.

To find the weighted average grade, follow these steps:

• Step 1: Convert the percentage weights into decimal form:
• 40% to 0.40
• 20% to 0.20
• 15% to 0.15
• 25% to 0.25
• Step 2: Calculate the contribution of each grade by multiplying it by its weight:
• Grade 68 with weight 0.40: $68 \times 0.40 = 27.2$
• Grade 73 with weight 0.20: $73 \times 0.20 = 14.6$
• Grade 94 with weight 0.15: $94 \times 0.15 = 14.1$
• Grade 91 with weight 0.25: $91 \times 0.25 = 22.75$
• Step 3: Sum the contributions:
$27.2 + 14.6 + 14.1 + 22.75 = 78.65$

Since the weights sum up to 100%, the weighted average grade is directly this sum.

Therefore, the student's weighted average grade is $78.65$.

To solve the weighted average, we will use the following formula:

exam 2 * weight of evaluation 2 + exam 1 * weight of the evaluation 1 = Weighted average

We will place the data in the formula, where the weights will be in decimal numbers:

0.3*79 + 0.7*83 = 
23.7+58.1 = 

81.8

In order to determine the hotel rating, we will calculate an average.

Let's remember the weighted average formula:

(value A X weight percentage A)+(value B X weight percentage B)...

First, let's add all the percentages together to make sure we reach 100 percent:

$45\%+30\%+10\%+10\%+5=100\%$

Now we'll multiply each factor by its weight percentage, convert the percentages to decimal numbers, and add them as follows:

$0.45\times3.5+0.3\times4+0.1\times2+0.1\times4.5+0.05\times1=$

Let's solve the multiplication problems first:

$1.575+1.2+0.2+0.45+0.05=$

We'll add them together and get: 3.475 and that's the hotel rating

To calculate Gabriel's final weighted average, we'll use the formula for a weighted average:

$\text{Weighted Average} = (\text{Test 1 Grade} \times \text{Weight 1}) + (\text{Exam Grade} \times \text{Weight 2}) + (\text{Test 2 Grade} \times \text{Weight 3}) + (\text{Project Grade} \times \text{Weight 4})$

Next, calculate the contribution of each component:

• Test 1 Contribution: $84 \times 0.30 = 25.2$
• Exam Contribution: $73 \times 0.15 = 10.95$
• Test 2 Contribution: $87 \times 0.35 = 30.45$
• Project Contribution: $92 \times 0.20 = 18.4$

Now, add all these contributions to find the final weighted average:

$\text{Weighted Average} = 25.2 + 10.95 + 30.45 + 18.4 = 85$

Therefore, Gabriel's final weighted average is $\boxed{85}$.

To determine if Ramiro's grades meet the average required, we calculate the weighted average of his grades. This involves applying the formula for calculating a weighted average:

• Multiply each grade by its corresponding weight:
  $92 \times 0.40 = 36.8$
  $78 \times 0.15 = 11.7$
  $98 \times 0.10 = 9.8$
  $83 \times 0.35 = 29.05$
• Sum these products to calculate the weighted average:
  $\text{Weighted Average} = 36.8 + 11.7 + 9.8 + 29.05 = 87.35$

Since 87.35 is greater than the required 85, Ramiro's weighted average meets the high school requirement.

Therefore, the solution to the problem is that Ramiro will be admitted to the high school. His weighted grade average is $87.35$.

To solve this problem, we'll compute the weighted average speed of the roller coaster using the given weights and speeds:

• Step 1: Convert the weights from percentages to decimal form:

  • 20% becomes 0.20

  • 15% becomes 0.15

  • 5% becomes 0.05

  • 35% becomes 0.35

  • 5% becomes 0.05

  • 20% becomes 0.20

• Step 2: Calculate each term by multiplying the speed by its corresponding weight:

• $\begin{aligned} 100 \times 0.20 &= 20, \\ 40 \times 0.15 &= 6, \\ 130 \times 0.05 &= 6.5, \\ 60 \times 0.35 &= 21, \\ 15 \times 0.05 &= 0.75, \\ 75 \times 0.20 &= 15. \end{aligned}$

• Step 3: Sum the results of these products:

• $20 + 6 + 6.5 + 21 + 0.75 + 15 = 69.25$

• Step 4: Since the weights sum to 1, the weighted average speed is:

Therefore, the average speed of the roller coaster is 69.25 km/h.

To solve this problem, we'll follow these steps:

• Step 1: Determine the number of students with no siblings.
• Step 2: Calculate the total number of siblings.
• Step 3: Use the weighted average formula to find the average number of siblings per student.

Step 1: Determine the number of students with no siblings.

The problem provides us with the following information:
- 20 students have 2 siblings.
- 15 students have 1 sibling.
- Total number of students = 45.

The number of students with no siblings can be calculated as:

Number of students with no siblings = Total number of students - Students with 2 siblings - Students with 1 sibling.

$45 - 20 - 15 = 10$

Step 2: Calculate the total number of siblings.

The total number of siblings is calculated by summing the product of the number of siblings and the number of students in each category:

Total number of siblings = (2 siblings × 20 students) + (1 sibling × 15 students) + (0 siblings × 10 students)

$(2 \times 20) + (1 \times 15) + (0 \times 10) = 40 + 15 + 0 = 55$

Step 3: Use the weighted average formula.

The weighted average for the number of siblings per student is given by:

Average number of siblings = $\frac{\text{Total number of siblings}}{\text{Total number of students}}$

$\frac{55}{45} = \frac{11}{9} \approx 1.22$

Therefore, the students in the class have, on average, $1.22$ siblings.

To solve this problem, we'll calculate the average number of defective items per day using weighted averages:

• Step 1: Compute the total number of defective items.
  • 3 days × 4 defective items/day = 12 defective items
  • 2 days × 7 defective items/day = 14 defective items
  • 1 day × 0 defective items/day = 0 defective items
  • 10 days × 18 defective items/day = 180 defective items
• Step 2: Sum all defective items calculated: $12 + 14 + 0 + 180 = 206 \text{ defective items}$
• Step 3: Compute the total number of days: $3 + 2 + 1 + 10 = 16 \text{ days}$
• Step 4: Calculate the average number of defective items per day using: $\frac{206}{16} = 12.875$

Therefore, the average number of defective items per day is $12.875$.

To solve this problem, we'll calculate the weighted average price per kilogram of the feather mix:

• First, determine the percentage of ostrich feathers. Since pigeon and goose feathers add up to 74%, ostrich feathers comprise 26% of the mix.
• Next, convert the percentages to decimals:
  • Pigeon feathers: $42\% = 0.42$
  • Goose feathers: $32\% = 0.32$
  • Ostrich feathers: $26\% = 0.26$
• Now, apply the weighted average formula: $\text{Weighted Average Price} = (35 \times 0.42) + (95 \times 0.32) + (72 \times 0.26)$ Calculating these:
  • Pigeon contribution: $14.7$
  • Goose contribution: $30.4$
  • Ostrich contribution: $18.72$
• Summing these contributions gives us the total weighted average price: $14.7 + 30.4 + 18.72 = 63.82$

Therefore, the price per kilogram of the feather mix is \63.82\).

To solve this problem, we'll follow these steps:

• Identify the contribution of the aluminum component to the overall cost.

• Identify the contribution of the iron component to the overall cost.

• Combine the contributions to find the total cost of the alloy.

Let's perform the calculations:

1. Aluminum's Contribution:
The alloy is 30% aluminum, which costs 5 per 100 grams. Hence, the contribution from aluminum is:
\( 0.30 \times 5 = 1.5 \text{ dollars}

2. Iron's Contribution:
The alloy is 70% iron, which costs 17 per 100 grams. Hence, the contribution from iron is:
\( 0.70 \times 17 = 11.9 \text{ dollars}

3. Total Cost of the Alloy:
To obtain the total cost of 100 grams of the alloy, we add the contributions from both components:
$1.5 + 11.9 = 13.4 \text{ dollars}$

Therefore, the price of 100 grams of the alloy is $13.4 \text{ dollars}$.

To solve this problem, we'll follow these steps:

• Step 1: Calculate the weighted score for each component.
• Step 2: Sum these scores and express the total weighted score needed for the average.
• Step 3: Solve for the unknown final exam grade using the weighted average formula.

Now, let's work through each step:

Step 1: Calculate the individual components' weighted scores:

• Attendance: $95 \times 0.10 = 9.5$
• Assessment: $82 \times 0.10 = 8.2$
• Assignments: $100 \times 0.20 = 20.0$

Step 2: Calculate the total weighted score needed to achieve an average of 92.

The weighted average formula is given by:

Simplifying what we know:

Step 3: Solve for the final exam grade.

First, isolate the weighted exam score:

Next, solve for the actual final exam grade:

Therefore, the grade the student received on the final exam is $90.5$.

To solve this problem, we need to calculate the arithmetic mean of the given dice rolls.

First, let's identify all the given numbers:
The results from the dice rolls are $4, 3, 1, 6, 5, 3, 1$.

We will follow these steps to find the average:

• Step 1: Sum all the numbers: $4 + 3 + 1 + 6 + 5 + 3 + 1$.
• Step 2: Count the total number of rolls: There are 7 numbers.
• Step 3: Divide the sum obtained in Step 1 by the count from Step 2 to determine the average.

Let's execute these steps:

Step 1: Calculate the sum of the numbers:
$4 + 3 + 1 + 6 + 5 + 3 + 1 = 23$.

Step 2: Count the number of rolls: There are $7$ rolls.

Step 3: Calculate the average:
$\text{Average} = \frac{23}{7} = 3.2857\ldots$

The average can be rounded to two decimal places, giving us approximately $3.29$.

Thus, the average number resulting from the dice rolls is $\mathbf{3.29}$.

To solve this problem, we'll proceed with the following steps:

• Step 1: Define the weights and scores for each exam.
• Step 2: Formulate the equation for the weighted average.
• Step 3: Solve the equation for $x$, the unknown score on the first exam.

Now, let's work through each step:

Step 1: We know:

• Weight of the first exam: $0.55$
• Weight of the second exam: $0.45$
• Score on the second exam: $76$
• Final average score: $84$
• Unknown score (first exam): Let it be $x$

Step 2: We use the weighted average formula:

Step 3: Now solve for $x$.

First, calculate the contribution of the second exam:

Substitute this back into the equation:

Subtract 34.2 from both sides to solve for $0.55 \times x$:

Finally, divide both sides by 0.55 to isolate $x$:

Calculate the result:

Therefore, the solution to the problem is $\mathbf{90.55}$.