# Computations Involving the Mean, Sample Size, and Sum of a Data Set Online Quiz

Following quiz provides Multiple Choice Questions (MCQs) related to Computations Involving the Mean, Sample Size, and Sum of a Data Set. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz. Q 1 - The average of four numbers is 12. If three of those numbers are 7, 9 and 15, find the last number.

### Explanation

Step 1:

Let the last number be = x

Average = $\frac{(7 + 9 + 15 + x)}{4}$ = 12.

Step 2:

Solving for x;

31 + x = 48; x = 48 – 31 = 17

Q 2 - The mean of a, b and c is 26. The sum of a and twice the value of c is 28. What is a + 2b equal to?

### Explanation

Step 1:

Given a + b + c = 26 and a + 2c = 28

Step 2:

Subtracting the first from second we get c – b = 28 -26 = 2; c = 2 + b

Step 3:

Plugging value of c in first equation we get a + 2b = 24

Q 3 - The average of x and 3 is equal to the average of x, 6 and 9. Find x.

### Explanation

Step 1:

Average of x and 3 = $\frac{(x + 3)}{2}$; Average of x, 6, and 9 = $\frac{(x + 6 + 9)}{3}$

Step 2:

Given $\frac{(x + 3)}{2} = \frac{(x + 15)}{3}$

Solving we get 3x + 9 = 2x + 30 or 3x – 2x = x = 30 – 9 = 21

Step 3:

So x = 21.

Q 4 - 5 consecutive integers have an average of 123. What's the average of the first three of those integers?

### Explanation

Step 1:

Let the consecutive integers be a -2, a -1, a, a + 1 and a + 2

Average $\frac{(a -2 + a -1 + a + a + 1 + a + 2)}{5} = \frac{5a}{5}$ = 123

Step 2:

The numbers are 121, 122, 123, 124 and 125

Average of first three numbers = 122

Q 5 - 5 consecutive odd integers have an average of 101. What's the average of the first three of those integers?

### Explanation

Step 1:

- Let the consecutive odd integers be a - 4, a -2, a, a + 2 and a + 4

Average $\frac{(a - 4 + a - 2 + a + a + 2 + a + 4)}{5} = \frac{5a}{5}$ = 101

Step 2:

The numbers are 97, 99, 101, 103 and 105

Average of first three numbers = 99

Q 6 - P is a set of numbers with a mean of 10. Q is a set of numbers found by adding 2 to each of the numbers in set P. What is the average of the numbers in set Q?

### Explanation

Step 1:

Let there be n numbers x1, x2, ….xn in set P. Their mean = 10.

Total in set P = (x1 + x2 + …xn) = 10n

Step 2:

In set Q, n numbers x1+2, x2+2 ….xn+2.

Their total (x1+2, x2+2 ….xn+2) = (x1 + x2 + …xn) + 2n = 10n + 2n = 12n

Step 3

Mean of set Q = $\frac{Total}{n}= \frac{12n}{n}$ = 12

Q 7 - The average temperature over a 10-day period was 85 degrees. For the first six days, the average temperature was 87 degrees. Over the next 4 days, what was the average temperature?

### Explanation

Step 1:

Total degrees over 10 day period = 10 × 85 = 850

Step 2:

Let average temperature over 4 days = x

87 × 6 + 4 x = 850

Solving for x, 4x = 850 – 522 = 328

So x = $\frac{328}{4}$ = 82 degrees.

Q 8 - The average of 14, 18 and 19 is 10 more than the average of 8 and what number?

### Explanation

Step 1:

Average of 14, 18 and 19 = $\frac{(14 + 18 + 19)}{3} = \frac{51}{3}$ = 17

Step 2:

Let required number by x

Average of x and 8 = $\frac{(8 + x)}{2}$

Given $\frac{(8 + x)}{2}$ + 10 = 17

So $\frac{(8 + x)}{2}$ = 17 – 10 = 7

8 + x = 14; x = 14 – 8 = 6.

Step 3

So required number = 6

Q 9 - 7 consecutive even integers have an average of 48. Find the average of the greatest two of those integers.

### Explanation

Step 1:

Let the consecutive even integers be x – 6, x – 4, x – 2, x, x + 2, x + 4, x + 6

Their average = $\frac{(x – 6 + x – 4 + x – 2 + x + x + 2 + x + 4 + x + 6)}{7} = \frac{7x}{7}$ = 48. So x = 48.

Step 2:

So the numbers are 42, 44, 46, 48, 50, 52, 54

The average of the two greatest of these integers 52 and 54 is $\frac{(52 + 54)}{2}=$ = 53

Q 10 - The average height of students in a school is 5 foot 6 inches. If the 200 girls at the school average 5 foot 4 inches, what is the average height of the 180 boys at the school?

### Explanation

Step 1:

Let the average height of the boys = x Then total height of 180 boys = 180x

Total height of 200 girls = 200 × $5\frac{4}{12} = \frac{3200}{3}$

Step 2:

Total height of boys and girls = 180x + $\frac{3200}{3}$

= Average x number of boys and girls = 5$\frac{6}{12}$ × 380

180x + $\frac{3200}{3}$ = 2090; 180x = $\frac{(6270 − 3200)}{3} = \frac{3070}{3}$; x = $\frac{3070}{540}$ = 5.69 feet.

computations_involving_mean_sample_size_and_sum_of_data_set.htm 