Measures of Central Tendency Test

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Measures of Central Tendency Test - 39

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Question 1

1 / -0

Marks	0-5	5-10	10-15	15-20	20-25	25-30	30-35	35-40	40-45	45-50
Frequency	3	5	7	8	10	11	14	19	15	13

For the following distribution, find the mean using step deviation method. (Round off your answer to the nearest whole number)

$$29$$

$$31$$

$$35$$

$$37$$

Solution

X	Mid point (x)	Frequency (F)	i=class interval width	A=22.5 Assumed mean	u=$$\dfrac{x-A}{i}$$	fu
0-5	2.5	3	5	22.5	-4	-12
5-10	7.5	5	5	22.5	-3	-15
10 -15	12.5	7	5	22.5	-2	-14
15-20	17.5	8	5	22.5	-1	-8
20-25	22.5=A	10	5	22.5	0	0
25-30	27.5	11	5	22.5	1	11
30-35	32.5	14	5	22.5	2	28
35-40	37.5	19	5	22. 5	3	57
40-45	42.5	15	5	22.5	4	60
45-50	47.5	13	5	22.5	5	65
		$$\Sigma f=105$$				$$\Sigma fu=172$$

The formula used for arithmetic mean of grouped data by step deviation method is, $$\overline {X} =A + \dfrac{\sum fu}{\sum f} \times i$$
$$A =$$ Assumed mean of the given data
$$\sum$$ = Summation of the frequencies given in the grouped data
$$\sum fu$$ = Summation of the frequencies and deviation of a given mean data
$$u= \dfrac{(x - A)}{i}$$
$$i =$$ Class interval width
$$\overline {X}$$ = arithmetic mean
$$\overline {X} =22.5 +\dfrac{172}{105}\times 5$$
$$= 22.5 + 8.19$$
$$= 30.69$$ $$\approx$$ $$31$$

Question 2

1 / -0

Marks	0-10	10-20	20-30	30-40	40-50	50-60	60-70	70- 80	80-90	90-100
Frequency	3	5	6	7	8	9	10	12	6	4

Find the mean mark using step deviation method:

$$54$$

$$55$$

$$56$$

$$57$$

Solution

X	Mid Point (X)	Frequency (F)	i=class interval width	A=45 Assumed mean	u=$$\dfrac{x-A}{i}$$	fu
0-10	5	3	10	45	-4	-12
10-20	15	5	10	45	-3	-15
20-30	25	6	10	45	-2	-12
30-40	35	7	10	45	-1	-7
40-50	45=A	8	10	45	0	0
50-60	55	9	10	45	1	9
60-70	65	10	10	45	2	20
70-80	75	12	10	45	3	36
80-90	85	6	10	45	4	24
90-100	95	4	10	45	5	20
		$$\Sigma f=70$$				$$\Sigma fu=63$$

The formula used for arithmetic mean of grouped data by step deviation method is, $$\overline {X} =A + \dfrac{\sum fu}{\sum f} \times i$$
$$A =$$ Assumed mean of the given data
$$\sum$$ = Summation of the frequencies given in the grouped data
$$\sum fu$$ = Summation of the frequencies and deviation of a given mean data
$$u = \dfrac{(x - A)}{i}$$
$$i =$$ Class interval width
$$\overline {X}$$ = arithmetic mean
$$\overline {X} =45 +\dfrac{63}{70}\times 10$$
$$= 45 + 9$$
$$= 54$$

Question 3
1 / -0

You grew sixty plants using special soil. You measure their lengths and the values are tabulated below.
Length(cm) $$120-125$$ $$125-130$$ $$130-135$$ $$135-140$$ $$140-145$$ $$145-150$$
Number of plants $$8$$ $$10$$ Y $$16$$ $$8$$ $$12$$
The mean length of the plants is $$136$$cm. Find y in the table.

A
$$10$$

B
$$8$$

C
$$6$$

D
$$4$$

Solution

Length(cm) $$(f_i)$$ $$x_i$$ $$f_ix_i$$
$$120-125$$ $$8$$   $$122.5$$    $$122.5\times 8=980$$
$$125-130$$ $$10$$   $$127.5$$   $$127.5\times 10=1275$$
$$130-135$$ Y   $$132.5$$ $$y\times 132.5=132.5y$$
$$135-140$$ $$16$$   $$137.5$$ $$137.5\times 16=2200$$
$$140-145$$ $$8$$   $$142.5$$   $$142.5\times 8=1140$$
$$145-150$$ $$12$$   $$147.5$$ $$147.5\times 12=1770$$

$$\Sigma f_i=60$$ $$\Sigma f_ix_i=7365+795y$$
Arithmetic mean using direct method formula is $$\overline{x}=\displaystyle\frac{\Sigma f_i x_i}{\Sigma f_i}$$

$$136 = (7365 + 795y)/60$$
$$136 \times 60 = 7365 + 132.5y$$
$$ 8160 - 7365 = 132.5y$$
$$y = \frac{795}{132.5}$$
$$y = 6$$
Therefore, the value of y is $$6$$.
Question 4
1 / -0

The ages of the $$115$$ people who live on an apartment are grouped as follows$$:$$ The mean length of the plants is $$33.43$$years using direct method. Find y in the table.
Age(years) Number of people
$$0-10$$ $$10$$
$$10-20$$ $$15$$
$$20-30$$ $$26$$
$$30-40$$ Y
$$40-50$$ $$23$$
$$50-60$$ $$16$$
$$60-70$$ $$3$$
$$70-80$$ $$1$$

A
$$23$$

B
$$28$$

C
$$15$$

D
$$21$$

Solution

Length(cm)  No. of plants$$(f_i)$$ Mid point $$x_i$$ $$f_ix_i$$
$$0-10$$   $$10$$ $$5$$ $$10\times 5=50$$
$$10-20$$  $$15$$ $$15$$ $$15\times 15=225$$
$$20-30$$  $$26$$ $$25$$ $$26\times 25=650$$
$$30-40$$  $$y$$ $$35$$ $$y\times 35=35y$$
$$40-50$$  $$23$$ $$45$$ $$23\times 45=1035$$
$$50-60$$  $$16$$ $$55$$ $$16\times 55=880$$
$$60-70$$ $$3$$ $$65$$ $$3\times 65=195$$
$$70-80$$  $$1$$ $$75$$ $$1\times 75=75$$
$$\Sigma f_i=115$$ $$\Sigma f_ix_=3110+35y$$

Arithmetic mean using direct method formula is
$$\overline{x}=\displaystyle\frac{\Sigma f_i x_i}{\Sigma f_i}$$

$$33.43 = (3110 + 35y)/115$$

$$433.43 \times 115 = 3110 + 35y$$

$$3844.45 - 3110 = 35y$$

$$y = 734.45/35$$

$$y = 20.9$$

Therefore, approximately the value of $$y$$ is $$21.$$

Question 5

1 / -0

Using step deviation method find the mean.

X	20-40	40-60	60-80	80-100
frequency	4	8	12	16

$$40$$

$$50$$

$$60$$

$$70$$

Solution

X	Mid point (X)	Frequency (F)	i=class interval width	A=50 Assumed Means	u=$$\dfrac{x-A}{i}$$	fu
20-40	30	4	20	50	-1	-4
40-60	50=A	8	20	50	0	0
60-80	70	12	20	50	1	12
80-100	90	16	20	50	2	32
		$$\Sigma f=40$$				$$\Sigma f u=40$$

The formula used for arithmetic mean of grouped data by step deviation method is, $$\overline {X} =A + \dfrac{\sum fu}{\sum f} \times i$$
$$A =$$ Assumed mean of the given data
$$\sum$$ = Summation of the frequencies given in the grouped data
$$\sum fu$$ = Summation of the frequencies and deviation of a given mean data
$$u = \frac{(x - A)}{i}$$
$$i =$$ Class interval width
$$\overline {X}$$ = arithmetic mean
$$\overline {X} =50 +\dfrac{40}{40}\times 20$$
$$= 50 + 20$$
$$= 70$$

Question 6
1 / -0

Find the measures of central tendency fro the data set $$2, 4, 5, 1, 7, 2, 3$$.

A
mean $$= 3.42$$, median $$= 3$$ and mode $$= 2$$

B
mean $$= 3$$, median $$= 3$$ and mode $$= 2$$

C
mean $$= 3.42$$, median $$= 3$$ and mode $$= 3$$

D
mean $$= 3$$, median $$= 3$$ and mode $$= 3$$

Solution

Mean, median and mode of a data set are the measures of central tendency.
Step 1: Mean $$=$$ $$\dfrac{\sum X}{n}$$
$$\sum X$$ $$=$$ sum of the data values.
$$n =$$ number of the data values
Mean $$=$$ $$\dfrac{2 + 4+5+1+7+2+3}{7}=3.42$$
Step 2: The data set in the ascending order is $$1, 2, 2, 3, 4, 5, 7 $$
So, Median of the set is $$3$$. [Median is the middle data value of the ordered set.]
Step 3: Mode is the data value(s) that appear most often in the data set. So, the modes of the data set are $$2$$.
So, the measures of central tendency of the given set of data are mean $$= 3.42$$, median $$= 3$$ and mode $$= 2$$.
Question 7
1 / -0

The measure of central tendency of mean is affected by

A
variables

B
distributions

C
outliers

D
range

Solution

The measure of Central tendency of mean is affected by $$outlers$$ ,
because outliers are observation that lies on abnormal distance from
other values in a random sample data.
Question 8
1 / -0

If the data set is perfectly normal, the mean, median and mode are _____to each other.

A
Equal

B
Unequal

C
Similar

D
Not comparable

Solution

Since , when data is Normal it will be in $$bell shape$$ and also
$$symmetrical$$. In that case $$Median$$ , $$Mean$$ & $$Mode$$ of data are $$Equal$$.
Question 9
1 / -0

When the mean is the best measure of central tendency?

A
continuous and symmetrical data

B
discontinuous and symmetrical data

C
skewed data only

D
continuous data only

Solution

$$\Rightarrow$$ The mean is usually the best measure of central tendency to use when your data distribution is continuous and symmetrical, such as when your data is normally distributed.
$$\Rightarrow$$ However, it all depends on what you are trying to show from your data.
$$\Rightarrow$$ The mean is equal to the sum of all the values in the data set divided by the number of values in the data set.
$$\overline{x}=\dfrac{x_1+x_2+x_3+...x_n}{n}$$
Question 10
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If the average (arithmetic mean) of $$t$$ and $$(t+2)$$ is $$x$$ and if the average of $$t$$ and $$(t-2)$$ is $$y$$, what is the average of $$x$$ and $$y$$?

A
$$1$$

B
$$\cfrac{t}{2}$$

C
$$t$$

D
$$t+\cfrac{1}{2}$$

Solution

Given, the average (arithmetic mean) of $$t$$ and $$t+2$$ is $$x$$ and the average of $$t$$ and $$ t−2$$ is $$y$$

Then $$x=\dfrac{t+(t+1)}{2}=\dfrac{2t+1}{2}$$

And $$y=$$ $$y=\dfrac{t+(t-1)}{2}=\dfrac{2t-1}{2}$$

Then average $$x$$ and $$y =$$ $$\dfrac{x+y}{2}=\dfrac{\frac{2t+1}{2}+\frac{2t-1}{2}}{2}=\dfrac{2t+1+2t-1}{2\times 2}=\dfrac{4t}{4}=t$$

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Length(cm)	$$120-125$$	$$125-130$$	$$130-135$$	$$135-140$$	$$140-145$$	$$145-150$$
Number of plants	$$8$$	$$10$$	Y	$$16$$	$$8$$	$$12$$

Age(years)	Number of people
$$0-10$$	$$10$$
$$10-20$$	$$15$$
$$20-30$$	$$26$$
$$30-40$$	Y
$$40-50$$	$$23$$
$$50-60$$	$$16$$
$$60-70$$	$$3$$
$$70-80$$	$$1$$

Measures of Central Tendency Test - 39

Result

Measures of Central Tendency Test - 39

Score

-

Rank

-

SHARING IS CARING

Question 1

$$29$$

$$31$$

$$35$$

$$37$$

Solution

Question 2

$$54$$

$$55$$

$$56$$

$$57$$

Solution

Question 3

$$10$$

$$8$$

$$6$$

$$4$$

Solution

Question 4

$$23$$

$$28$$

$$15$$

$$21$$

Solution

Question 5

$$40$$

$$50$$

$$60$$

$$70$$

Solution

Question 6

mean $$= 3.42$$, median $$= 3$$ and mode $$= 2$$

mean $$= 3$$, median $$= 3$$ and mode $$= 2$$

mean $$= 3.42$$, median $$= 3$$ and mode $$= 3$$

mean $$= 3$$, median $$= 3$$ and mode $$= 3$$

Solution

Question 7

variables

distributions

outliers

range

Solution

Question 8

Equal

Unequal

Similar

Not comparable

Solution

Question 9

continuous and symmetrical data

discontinuous and symmetrical data

skewed data only

continuous data only

Solution

Question 10

$$1$$

$$\cfrac{t}{2}$$

$$t$$

$$t+\cfrac{1}{2}$$

Solution

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