Statistics Test

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Statistics Test - 49

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Weekly Quiz Competition

Question 1
1 / -0

The table is a frequency table for number of days and temperature obtained in a month. Find the median day of temperature.
Number of days
$$3$$
$$4$$
$$5$$
$$6$$
$$7$$
Temperature
$$^{o}C$$
$$20$$
$$30$$
$$10$$
$$15$$
$$5$$

A
$$3$$

B
$$4$$

C
$$5$$

D
$$6$$

Solution

$$Number\,of\,days$$ $$Temperature$$
$$^oC$$ $$c.f$$
$$3$$ $$20$$ $$20$$
$$4$$ $$30$$ $$50$$
$$5$$ $$10$$ $$60$$
$$6$$ $$15$$ $$75$$
$$7$$ $$5$$ $$80$$
Here, $$N=80$$ then, $$\dfrac{N}{2}=\dfrac{80}{2}=40$$

We find that the cummulative frequency greater than $$\displaystyle\frac{N}{2}(40)$$ is $$50$$ and value of $$x$$ corresponding to $$50$$ is $$4$$.
$$\therefore$$ Median day of temperature $$=4$$
Question 2
1 / -0

The frequency table displays the number of students of a class who spend varying amounts of time for doing their homework. Find the median of time spend doing their home work.
time (in sec)
$$2$$
$$4$$
$$6$$
$$8$$
$$10$$
$$12$$
$$14$$
frequency
$$1$$
$$2$$
$$3$$
$$2$$
$$1$$
$$4$$
$$5$$

A
$$10$$

B
$$12$$

C
$$11$$

D
$$14$$

Solution

$$Time\,(in\,sec)$$ $$Frequency$$ $$c.f$$
$$2$$ $$1$$ $$1$$
$$4$$ $$2$$ $$3$$
$$6$$ $$3$$ $$6$$
$$8$$ $$2$$ $$8$$
$$10$$ $$1$$ $$9$$
$$12$$ $$4$$ $$13$$
$$14$$ $$5$$ $$18$$
$$\Rightarrow$$ $$N=18$$ (even)
$$Median=\dfrac{1}{2}\left[\left(\dfrac{18}{2}\right)^{th}obs+\left(\dfrac{18}{2}+1\right)^{th}obs\right]$$

$$=\dfrac{1}{2}\left[9^{th}obs+10^{th}obs\right]$$
Here, $$9^{th}obs=10$$ and $$11^{th}obs=12$$
$$\therefore$$ Required median $$=\dfrac{1}{2}[10+12]$$
$$=11$$

Question 3

1 / -0

The frequency table shows the number of chocolates that were distributed among the students of different grades. Find the median number of grade students.

Grade	$$6$$	$$7$$	$$8$$
Number of chocolates distributed	$$10$$	$$24$$	$$11$$

$$6$$

$$7$$

$$8$$

$$5$$

Solution

Grade	Number of chocolates	Position
$$6$$	$$10$$	$$10$$
$$7$$	$$24$$	$$34$$
$$8$$	$$11$$	$$45$$

Sum of number of chocolates $$=10+21+11\\ =45(odd\quad number)\quad $$

Median at $$\left( \cfrac { n+1 }{ 2 } \right) position=\left( \cfrac { 45+1 }{ 2 } \right) =23$$

$$23$$rd position comes after $$10$$th but before $$34$$th position

So, $$Median=7$$

Question 4

1 / -0

What is the median for the following data?

shoe size	$$2$$	$$3$$	$$4$$	$$5$$	$$6$$
frequency	$$10$$	$$20$$	$$12$$	$$14$$	$$11$$

$$4$$

$$5$$

$$6$$

$$2$$

Solution

Shoe Size	Frequency	Position(cf)
$$2$$	$$10$$	$$10$$
$$3$$	$$20$$	$$30$$
$$4$$	$$12$$	$$42$$
$$5$$	$$14$$	$$56$$
$$6$$	$$11$$	$$67$$

Sum of frequency $$=10+20+12+147+11\\ =67(odd)$$

Median is at $$\quad \quad \left( \cfrac { n+1 }{ 2 } \right) \\ =\quad \left( \cfrac { 67+1 }{ 2 } \right) =\cfrac { 68 }{ 2 } =34$$

$$3rd$$ posotion comes after $$30th$$ but before $$42$$

$$Median=4$$

Question 5

1 / -0

The frequency table is shown. Find the median for the following data:

Number of hours	$$2$$	$$4$$	$$6$$	$$8$$
Frequency	$$13$$	$$15$$	$$24$$	$$11$$

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Number of hours	Frequency	C.f
$$2$$	$$13$$	$$13$$
$$4$$	$$15$$	$$28$$
$$6$$	$$24$$	$$52$$
$$8$$	$$11$$	$$63$$

Sum of frequency $$=63(odd number)$$

Median is $$\quad \quad \left( \cfrac { n+1 }{ 2 } \right) position$$

Median is at $$=\quad \left( \cfrac { 63+1 }{ 2 } \right) =\cfrac { 64 }{ 2 } =32$$

$$32$$ is greater than $$28$$ but less than $$52$$

Median is $$6Ans$$

Question 6

1 / -0

Evangeline made a frequency table of the scores of her students in a test. Find the median score.

Score	$$12$$	$$24$$	$$36$$	$$48$$
Frequency	$$2$$	$$3$$	$$4$$	$$1$$

$$12$$

$$24$$

$$30$$

$$36$$

Solution

Score	Frequency	C.f
$$12$$	$$2$$	$$2$$
$$24$$	$$3$$	$$5$$
$$36$$	$$4$$	$$9$$
$$48$$	$$1$$	$$10$$

Sum of frequency $$=2+3+4+1=10$$

Sum of frequency is even, the Median is at the average of $$\left( \cfrac { n }{ 2 } \right) \& \left( \cfrac { n }{ 2 } +1 \right) position$$

$$\left( \cfrac { n }{ 2 } \right) =\cfrac { 10 }{ 2 } \quad =5\\ \cfrac { n }{ 2 } +1=\cfrac { 10 }{ 2 } +1=6\quad \quad \quad $$

Average score of $$5$$th & $$6$$th $$=\cfrac { 24+36 }{ 2 } $$

$$=>\cfrac { 60 }{ 2 } =30$$

Question 7

1 / -0

The table shows the number of books on each number of subjects. Find the median.

Subject	$$2$$	$$3$$	$$5$$	$$6$$	$$3$$	$$9$$	$$10$$
number of books	$$20$$	$$39$$	$$10$$	$$50$$	$$20$$	$$50$$	$$30$$

$$2$$

$$3$$

$$5$$

$$6$$

Solution

Subject	Number of books	Cumulative frequency
$$2$$	$$20$$	$$20$$
$$3$$	$$39$$	$$59$$
$$5$$	$$10$$	$$69$$
$$6$$	$$580$$	$$119$$
$$3$$	$$20$$	$$139$$
$$9$$	$$50$$	$$189$$
$$10$$	$$30$$	$$219$$

Sum of number of books $$=219(odd)$$

Number of books is odd. So Median is at $$\quad \quad \left( \cfrac { n+1 }{ 2 } \right) position$$

$$=\quad \left( \cfrac { 219+1 }{ 2 } \right) =\cfrac { 220 }{ 2 } =110$$

$$110$$ is greater tha $$69$$ but less than $$119$$

Median $$=6$$

Question 8
1 / -0

A student noted that the number of cars passing through the spot on the road for $$200$$ periods each of $$10$$ minutes and summarized in a table given below. Find the mode of the data.
Number of cars
5-10
10-15
15-20
20-25
25-30
30-35
35-40
frequency
12
24
15
10
60
55
25

A
15.5

B
20.50

C
25.45

D
19.4

Solution

Here frequency of class interval $$25-30$$ is maximum.
So, it is the modal class
Now, $$l =$$ lower limit of modal class $$= 25$$
$$f_1$$ $$=$$ frequency of modal class $$= 60$$
$$f_0$$ $$=$$ frequency of class preceding the modal class $$= 10$$
$$f_2$$ $$=$$ frequency of class succeeding the modal class $$= 55$$
$$h =$$ class interval width $$= 5$$
So, Mode $$=$$ $$l +\left(\dfrac{f_{1}-f_{2}}{2f_{1}-f_{0}-f_{2}}\times h \right)$$
$$=$$ $$25 +\left(\dfrac{60-55}{2\times 60-10-55}\times 5 \right)$$
$$= 25 + 0.4545$$
$$= 25.45$$
Mode $$= 25.45$$
Question 9
1 / -0

What is the median class for the following data?

A
$$50.5-60.5$$

B
$$60.5-70.5$$

C
$$61-70$$

D
$$71-80$$

Solution

To find the median class:
First add the total number of frequencies.
Here, $$f = 100$$
$$=$$ $$\dfrac{Total \space\ number\space\ of \space\ frequency }{2}$$
$$=$$ $$\dfrac{100}{2} =50$$
Hence, the value 50 lies in the class interval 60.5-70.5 from the cummulative frequency table.
Question 10
1 / -0

Class interval
2-4
4-6
6-8
8-10
10-12
frequency
2
4
6
10
5
What is the mode for the grouped data?

A
$$10.5$$

B
$$12.5$$

C
$$13.7$$

D
$$9.1$$

Solution

Here frequency of class interval $$8-10$$ is maximum.
So, it is the modal class
Now, $$l =$$ lower limit of modal class $$= 8$$
$$f_1$$ $$=$$ frequency of modal class $$= 10$$
$$f_0$$ $$=$$ frequency of class preceding the modal class $$= 6$$
$$f_2$$ $$=$$ frequency of class succeeding the modal class $$= 5$$
$$h =$$ class interval width $$= 2$$
So, Mode $$=$$ $$l +\left(\dfrac{f_{1}-f_{2}}{2f_{1}-f_{0}-f_{2}}\times h \right)$$
$$=$$ $$8 +\left(\dfrac{10-5}{2\times 10-6-5}\times 2 \right)$$
$$= 8 + 1.1111$$
Mode $$= 9.1$$

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Re-Attempt Weekly Quiz Competition

Number of days	$$3$$	$$4$$	$$5$$	$$6$$	$$7$$
Temperature $$^{o}C$$	$$20$$	$$30$$	$$10$$	$$15$$	$$5$$

$$Number\,of\,days$$	$$Temperature$$ $$^oC$$	$$c.f$$
$$3$$	$$20$$	$$20$$
$$4$$	$$30$$	$$50$$
$$5$$	$$10$$	$$60$$
$$6$$	$$15$$	$$75$$
$$7$$	$$5$$	$$80$$

time (in sec)	$$2$$	$$4$$	$$6$$	$$8$$	$$10$$	$$12$$	$$14$$
frequency	$$1$$	$$2$$	$$3$$	$$2$$	$$1$$	$$4$$	$$5$$

$$Time\,(in\,sec)$$	$$Frequency$$	$$c.f$$
$$2$$	$$1$$	$$1$$
$$4$$	$$2$$	$$3$$
$$6$$	$$3$$	$$6$$
$$8$$	$$2$$	$$8$$
$$10$$	$$1$$	$$9$$
$$12$$	$$4$$	$$13$$
$$14$$	$$5$$	$$18$$

Number of cars	5-10	10-15	15-20	20-25	25-30	30-35	35-40
frequency	12	24	15	10	60	55	25

Class interval	2-4	4-6	6-8	8-10	10-12
frequency	2	4	6	10	5

Statistics Test - 49

Result

Statistics Test - 49

Score

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Rank

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SHARING IS CARING

Question 1

$$3$$

$$4$$

$$5$$

$$6$$

Solution

Question 2

$$10$$

$$12$$

$$11$$

$$14$$

Solution

Question 3

$$6$$

$$7$$

$$8$$

$$5$$

Solution

Question 4

$$4$$

$$5$$

$$6$$

$$2$$

Solution

Question 5

$$2$$

$$4$$

$$6$$

$$8$$

Solution

Question 6

$$12$$

$$24$$

$$30$$

$$36$$

Solution

Question 7

$$2$$

$$3$$

$$5$$

$$6$$

Solution

Question 8

15.5

20.50

25.45

19.4

Solution

Question 9

$$50.5-60.5$$

$$60.5-70.5$$

$$61-70$$

$$71-80$$

Solution

Question 10

$$10.5$$

$$12.5$$

$$13.7$$

$$9.1$$

Solution

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