Measures of Central Tendency Test

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Measures of Central Tendency Test - 40

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Question 1
1 / -0

Priya jogged for a total of $$30$$ minutes. Her average speed for the first $$10$$ minutes was $$5$$ miles per hour. During the last 20 minutes, she jogged $$2.5$$ miles. What was the average speed of her entire jog?

A
$$6\dfrac {1}{2} mph$$

B
$$3\dfrac {1}{3} mph$$

C
$$7\dfrac {1}{2} mph$$

D
$$7\dfrac {1}{3} mph$$

E
$$7\dfrac {2}{3} mph$$

Solution

Distance traveled in first $$10$$ minutes is $$5 \times \dfrac 16 = \cfrac 56$$

Distance traveled in next $$20$$ minutes is $$2.5 \times \dfrac 13 = \dfrac 56$$

Total distance traveled is $$\dfrac 56 + \dfrac 56 = \dfrac 53$$

Average speed is $$\dfrac { \frac 53 }{ \frac 12 } = \dfrac {10}{3}$$
Question 2
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The average (arithmetic mean) of all the grades on a certain algebra test was $$90$$. If the average of the $$8$$ males grades was $$87$$, and the average of the females grades was $$92$$, calculate the number of female students.

A
8

B
9

C
10

D
11

E
12

Solution

Let the female students be $$x$$ and male students is $$8$$, so the total number of students is $$x+8$$
$$\Rightarrow 90(x+8)=87 \times 8 + 92x$$
$$\Rightarrow 2x = 24$$
$$\Rightarrow x = 12$$
Question 3
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Find the arithmetic mean of $$\dfrac {1}{2}, \dfrac {1}{3}, 2n$$ and $$m$$.

A
$$\dfrac {5 + 12n + 6m}{24}$$

B
$$\dfrac {5 + 8n + 4m}{24}$$

C
$$\dfrac {5 + 2n + m}{4}$$

D
$$\dfrac {5 + 12n + 6m}{6}$$

E
$$\dfrac {5 + 2n + m}{12}$$

Solution

The arithmetic mean of $$\dfrac {1}{2}, \dfrac {1}{3}, 2n, m$$ is
$$\dfrac { \dfrac { 1 }{ 2 } +\dfrac { 1 }{ 3 } +2n+m }{ 4 } $$
$$=\dfrac { \dfrac { 5 }{ 6 } +2n+m }{ 4 } $$
$$=\dfrac { 5+12n+6m }{ 24 } $$
Question 4
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$$M$$ is a set consisting of a finite number of consecutive integers. If the median of the numbers in set $$M$$ is equal to one of the numbers in set $$M$$, which of the following must be true?
$$I.$$ The average (arithmetic mean) of the numbers in set $$M$$ equals the median.
$$II.$$ The number of numbers in set $$M$$ is odd.
$$III.$$ The sum of the smallest number and the largest number in set $$M$$ is even

A
None of these

B
$$II$$ only

C
$$I$$ and $$II$$ only

D
$$I$$ and $$III$$ only

E
$$I, II$$ and $$III$$
Question 5
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What is the weighted arithmetic mean of the first $$n$$ natural numbers whose weights are equal to the corresponding numbers?

A
$$\dfrac { 1 }{ 2 } \left( n+1 \right) $$

B
$$\dfrac { 1 }{ 2 } \left( n+2 \right) $$

C
$$\dfrac { 1 }{ 3 } \left( n+1 \right) $$

D
$$\dfrac { 1 }{ 3 } \left( 2n+1 \right) $$

Solution

Weight Arithmetic mean $$= \cfrac { 1*1+2*2+3*3+....{ n }^{ 2 } }{ 1+2+3+....n } $$
$$=\cfrac { n(2n+1)(n+1) }{ 6*\cfrac { n(n+1) }{ 2 } } \\ =\cfrac { 2n+1 }{ 3 } \\ =\cfrac { 1 }{ 3 } (2n+1)$$.
Question 6
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A sample of $$25$$ scores has a mean $$75$$, median $$79$$ and standard deviation of $$8$$. If you increase every score by $$10$$, which of the following is true?
$$I$$. The new mean is $$85$$
$$II$$. The new median is $$89$$
$$III$$. The new standard deviation is $$18$$

A
$$I$$ only

B
$$II$$ only

C
$$I$$ and $$II$$ only

D
none

E
$$I, II$$ and $$III$$

Solution

New Mean = Old Mean +$$10=75+10=85$$
New standard deviation= old standard deviation $$=8$$
New median = Old median$$+10=79+10=89$$
Question 7
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Brian got grades of $$92,89$$ and $$86$$ on his first three math tests. What grade must he get on his final test to have an overall average of $$90$$?

A
90

B
93

C
87

D
85

Solution

Let the grade he got on final test be $$x$$
The average of all four grades is $$90$$.
So, we have $$\dfrac {(92+89+86+x)}{4}=90$$
$$\Rightarrow x = 360-267 = 93$$
Question 8
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The sum of four consecutive odd integers $$w, x, y,$$ and $$z$$ is $$24$$. What is the median of the set $$\left \{w, x, y, z, 24\right \}$$?

A
$$3$$

B
$$5$$

C
$$7$$

D
$$9$$

E
$$24$$

Solution

Given that the sum of four consecutive odd integers is $$24$$.
which implies $$x=w+2 , y=w+4 , z=w+6$$.
Therefore $$4w+12=24$$ , which implies $$w=3$$.
So, the median of set $${3,5,7,9,24}$$ is $$7$$.
Question 9
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X, Y, Z are three sets of values.
Which of the following statements is true?

A
Mean of X is equal to Mode of Z.

B
Mean of Z is equal to Mode of Y.

C
Median of Y is equal to Mode of X.

D
Mean, Median and Mode of X are same.

Solution

Arranging the data ascending order we get
$$X:1,2,2,3,3,3,7$$; $$Y:3,5,5,7,8,9,12$$ and $$Z:2,3,4,4,4,7,11$$
Mean of $$X=\cfrac{1+2+2+3+3+3+7}{7}=3$$
Mean of $$Y=\cfrac{3+5+5+7+8+9+12}{7}=7$$
Mean of $$Z=\cfrac{2+3+4+4+47+11}{7}=5$$
Mode of $$X=3$$
Mode of $$Y=5$$
Mode of $$Z=4$$
Median of $$X=3$$
Median of $$Y=7$$
Median of $$Z=4$$
Question 10
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The table below shows the number of cars Jing sold each month last year. What is the median of the data in the table?

A
$$13$$

B
$$16$$

C
$$19$$

D
$$20.5$$

E
$$23.5$$

Solution

Number of cars sold are $$25,15,22,19,16,13,19,25,26,27,28,29$$
In ascending order, $$13,15,16,19,19,22,25,25,26,27,28,29$$
Middle most values are $$22$$ and $$25$$
Median is a middle most value or average of two middle most values.
$$\Rightarrow M=\dfrac {(22+25)}{2}=23.5$$

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Re-Attempt Weekly Quiz Competition

Measures of Central Tendency Test - 40

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Measures of Central Tendency Test - 40

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Question 1

$$6\dfrac {1}{2} mph$$

$$3\dfrac {1}{3} mph$$

$$7\dfrac {1}{2} mph$$

$$7\dfrac {1}{3} mph$$

$$7\dfrac {2}{3} mph$$

Solution

Question 2

8

9

10

11

12

Solution

Question 3

$$\dfrac {5 + 12n + 6m}{24}$$

$$\dfrac {5 + 8n + 4m}{24}$$

$$\dfrac {5 + 2n + m}{4}$$

$$\dfrac {5 + 12n + 6m}{6}$$

$$\dfrac {5 + 2n + m}{12}$$

Solution

Question 4

None of these

$$II$$ only

$$I$$ and $$II$$ only

$$I$$ and $$III$$ only

$$I, II$$ and $$III$$

Question 5

$$\dfrac { 1 }{ 2 } \left( n+1 \right) $$

$$\dfrac { 1 }{ 2 } \left( n+2 \right) $$

$$\dfrac { 1 }{ 3 } \left( n+1 \right) $$

$$\dfrac { 1 }{ 3 } \left( 2n+1 \right) $$

Solution

Question 6

$$I$$ only

$$II$$ only

$$I$$ and $$II$$ only

none

$$I, II$$ and $$III$$

Solution

Question 7

90

93

87

85

Solution

Question 8

$$3$$

$$5$$

$$7$$

$$9$$

$$24$$

Solution

Question 9

Mean of X is equal to Mode of Z.

Mean of Z is equal to Mode of Y.

Median of Y is equal to Mode of X.

Mean, Median and Mode of X are same.

Solution

Question 10

$$13$$

$$16$$

$$19$$

$$20.5$$

$$23.5$$

Solution

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