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Statistics - Mean, Median, Mode

For COMPETITION
Number of Total Problems: 9.
FOR PRINT ::: (Book)

Problem Num : 1

From : AMC10

Type:

Section:Statistics

Theme:
Adjustment# : 0

Difficulty: 1

'
An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?
$extbf{(A)} frac{31}{16}qquad extbf{(B)} 2qquad extbf{(C)} frac{17}{8}qquad extbf{(D)} 3qquad extbf{(E)} fra...$
'

Category Mean, Median, Mode

Analysis

Solution/Answer
The minimum and maximum can be achieved with the orders $5, 4, 3, 2, 1$ and $1, 2, 3, 4, 5$ .
$5,4,3,2,1 Rightarrow frac92,3,2,1 Rightarrow frac{15}{4},2,1 Rightarrow frac{23}{8},1 Rightarrow frac{31}{16}$
$1,2,3,4,5 Rightarrow frac32,3,4,5 Rightarrow frac94,4,5 Rightarrow frac{25}{8},5 Rightarrow frac{65}{16}$
The difference between the two is $frac{65}{16}-frac{31}{16}=frac{34}{16}=oxed{ extbf{(C)} frac{17}{8}}$ .

Answer:

Problem Num : 2

From : AMC10B

Type:

Section:Statistics

Theme:
Adjustment# : 0

Difficulty: 1

'
The median of the list $n, n + 3, n + 4, n + 5, n + 6, n + 8, n + 10, n + 12, n + 15$ is $10$ . What is the mean?
$extbf{(A) }4qquad extbf{(B) }6qquad extbf{(C) }7qquad extbf{(D) }10qquad extbf{(E) }11$
'

Category Mean, Median, Mode

Analysis

Solution/Answer
The median of the list is $10$ , and there are $9$ numbers in the list, so the median must be the 5th number from the left, which is $n+6$ .
We substitute the median for $10$ and the equation becomes $n+6=10$ .
Subtract both sides by 6 and we get $n=4$ .
$n+n+3+n+4+n+5+n+6+n+8+n+10+n+12+n+15=9n+63$ .
The mean of those numbers is $frac{9n-63}{9}$ which is $n+7$ .
Substitute $n$ for $4$ and $4+7=oxed{ extbf{(E) }11}$ .

Answer:

Problem Num : 3

From : AMC10

Type:

Section:Statistics

Theme:
Adjustment# : 0

Difficulty: 1

'
The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$ext{(A) }11qquad ext{(B) }12qquad ext{(C) }13qquad ext{(D) }14qquad ext{(E) }15$
'

Category Mean, Median, Mode

Analysis

Solution/Answer
As the unique mode is $8$ , there are at least two $8$ s.
As the range is $8$ and one of the numbers is $8$ , the largest one can be at most $16$ .
If the largest one is $16$ , then the smallest one is $8$ , and thus the mean is strictly larger than $8$ , which is a contradiction.
If the largest one is $15$ , then the smallest one is $7$ . This means that we already know four of the values: $8$ , $8$ , $7$ , $15$ . Since the mean of all the numbers is $8$ , their sum must be $64$ . Thus the sum of the missing four numbers is $64-8-8-7-15=26$ . But if $7$ is the smallest number, then the sum of the missing numbers must be at least $4cdot 7=28$ , which is again a contradiction.
If the largest number is $14$ , we can easily find the solution $(6,6,6,8,8,8,8,14)$ . Hence, our answer is $oxed{ ext{(D)} 14 }$ .

Note

The solution for $14$ is, in fact, unique. As the median must be $8$ , this means that both the $4^ ext{th}$ and the $5^ ext{th}$ number, when ordered by size, must be $8$ s. This gives the partial solution $(6,a,b,8,8,c,d,14)$ . For the mean to be $8$ each missing variable must be replaced by the smallest allowed value.

Answer:

Problem Num : 4

From : AMC8

Type:

Section:Statistics

Theme:
Adjustment# : 0

Difficulty: 1

'
The mean of a set of five different positive integers is 15. The median is 18. The maximum possible value of the largest of these five integers is
$ext{(A)} 19 qquad ext{(B)} 24 qquad ext{(C)} 32 qquad ext{(D)} 35 qquad ext{(E)} 40$
'

Category Mean, Median, Mode

Analysis

Solution/Answer
Since there is an odd number of terms, the median is the number in the middle, specifically, the third largest number is $18$ , and there are $2$ numbers less than $18$ and $2$ numbers greater than $18$ . The sum of these integers is $5(15)=75$ , since the mean is $15$ . To make the largest possible number with a given sum, the other numbers must be as small as possible. The two numbers less than $18$ must be positive and distinct, so the smallest possible numbers for these are $1$ and $2$ . One of the numbers also needs to be as small as possible, so it must be $19$ . This means that the remaining number, the maximum possible value for a number in the set, is $75-1-2-18-19=35, oxed{ ext{D}}$ .

Answer:

Problem Num : 5

From : AMC10

Type:

Section:Statistics

Theme:
Adjustment# : 0

Difficulty: 1

'
The average (mean) of $20$ numbers is $30$ , and the average of $30$ other numbers is $20$ . What is the average of all $50$ numbers?
$mathrm{(A) } 23qquad mathrm{(B) } 24qquad mathrm{(C) } 25qquad mathrm{(D) } 26qquad mathrm{(E) } 27$
'

Category Mean, Median, Mode

Analysis

Solution/Answer
Since the average of the first $20$ numbers is $30$ , their sum is $20cdot30=600$ .
Since the average of $30$ other numbers is $20$ , their sum is $30cdot20=600$ .
So the sum of all $50$ numbers is $600+600=1200$
Therefore, the average of all $50$ numbers is $frac{1200}{50}=24 Longrightarrow mathrm{(B)}$

Answer:

Problem Num : 6

From : AMC10

Type:

Section:Statistics

Theme:
Adjustment# : 0

Difficulty: 1

'
The Dunbar family consists of a mother, a father, and some children. The average age of the members of the family is $20$ , the father is $48$ years old, and the average age of the mother and children is $16$ . How many children are in the family?
$ext{(A)} 2 qquad ext{(B)} 3 qquad ext{(C)} 4 qquad ext{(D)} 5 qquad ext{(E)} 6$
'

Category Mean, Median, Mode

Analysis

Solution/Answer
Let $n$ be the number of children. Then the total ages of the family is $48 + 16(n+1)$ , and the total number of people in the family is $n+2$ . So
$20 = frac{48 + 16(n+1)}{n+2} Longrightarrow 20n + 40 = 16n + 64 Longrightarrow n = 6 mathrm{(E)}.$

Answer:

Problem Num : 7

From : NCTM

Type: Understanding

Section:Statistics

Theme:Integer Property
Adjustment# : 0

Difficulty: 2

Category Mean, Median, Mode

Analysis

Solution/Answer

Problem Num : 8

From : AMC10B

Type:

Section:Statistics

Theme:
Adjustment# : 0

Difficulty: 1

'
The average of the numbers $1, 2, 3,cdots, 98, 99,$ and $x$ is $100x$ . What is $x$ ?
$extbf{(A)} dfrac{49}{101} qquad extbf{(B)} dfrac{50}{101} qquad extbf{(C)} dfrac{1}{2} qquad extbf{(D)} df...$
'

Category Mean, Median, Mode

Analysis

Solution/Answer
We must find the average of the numbers from $1$ to $99$ and $x$ in terms of $x$ . The sum of all these terms is $frac{99(100)}{2}+x=99(50)+x$ . We must divide this by the total number of terms, which is $100$ . We get: $frac{99(50)+x}{100}$ . This is equal to $100x$ , as stated in the problem. We have: $frac{99(50)+x}{100}=100x$ . We can now cross multiply. This gives:
$egin{align*}100(100x)&=99(50)+x\10000x&=99(50)+x\9999x&=99(50)\101x&=50\x&=oxed{ extbf{(B)} fr...$

Answer:

Problem Num : 9

From : AMC10

Type:

Section:Statistics

Theme:
Adjustment# : 0

Difficulty: 1

'
An iterative average of the numbers 1, 2, 3, 4, and 5 is computed the following way. Arrange the five numbers in some order. Find the mean of the first two numbers, then find the mean of that with the third number, then the mean of that with the fourth number, and finally the mean of that with the fifth number. What is the difference between the largest and smallest possible values that can be obtained using this procedure?
$extbf{(A)} frac{31}{16}qquad extbf{(B)} 2qquad extbf{(C)} frac{17}{8}qquad extbf{(D)} 3qquad extbf{(E)} fra...$
'

Category Mean, Median, Mode

Analysis

Solution/Answer
The minimum and maximum can be achieved with the orders $5, 4, 3, 2, 1$ and $1, 2, 3, 4, 5$ .
$5,4,3,2,1 Rightarrow frac92,3,2,1 Rightarrow frac{15}{4},2,1 Rightarrow frac{23}{8},1 Rightarrow frac{31}{16}$
$1,2,3,4,5 Rightarrow frac32,3,4,5 Rightarrow frac94,4,5 Rightarrow frac{25}{8},5 Rightarrow frac{65}{16}$
The difference between the two is $frac{65}{16}-frac{31}{16}=frac{34}{16}=oxed{ extbf{(C)} frac{17}{8}}$ .

Answer:

Array ( [0] => 7936 [1] => 8228 [2] => 7765 [3] => 8300 [4] => 7812 [5] => 7852 [6] => 3007 [7] => 8135 [8] => 7932 ) 9