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## Karnataka 1st PUC Economics Question Bank Chapter 5 Measures of Central Tendency

### 1st PUC Economics Measures of Central Tendency TextBook Questions and Answers

Question 1.

Which average would be suitable in the following cases?

- Average size of readymade garments.
- Average intelligence of students in a class.
- Average production in a factory per shift.
- Average wages in an industrial concern.
- When the sum of absolute deviations from average is least.
- When quantities of the variable are in ratios.
- In case of open ended frequency distribution.

Answer:

- Mode
- Mode
- Mean
- Median
- Mean
- Weighted mean
- Median.

Question 2.

Indicate the most appropriate alternative from the multiple choices provided against each question:

i) The most suitable average for qualitative measurement is:

- Arithmetic mean
- Median
- Mode
- Geometric mean
- None of the above.

ii) Which average is affect most by the presence of extreme items?

- Median
- Mode
- Arithmetic Mean
- Harmonic Mean

iii) The algebraic sum of deviation of a set of n values from A.M. is

- N
- 0
- None of the above.

Answer:

QUESTIONS | APPROPRIATE ANSWER |

i) The most suitable average for qualitative measurement is | 3. Mode |

ii) Which average is affected most by the Presence of extreme items | 3. Arithmetic mean |

iii) The algebraic sum of deviation of a set of n value from A. M. is | 2. 0 (zero) |

Question 3.

Comment on whether the following statements are True or False

- The sum of deviation of items from median is zero.
- An average alone is not enough to compare series
- Arithmetic mean is a positional value
- The upper quartile is the lowest value of top 25% of items.
- Median is unduly affected by extreme observations.

Answer:

- False
- True
- False
- True
- False.

Question 4.

If the arithmetic mean of the data given below is 28, find

- The missing frequency and
- The median of the series.

Profit per retail shop (in Rs.) | Number of retail shops |

0- 10 | 12 |

10-20 | 18 |

20-30 | 27 |

30-40 | – |

40-50 | 17 |

50-60 | 6 |

Answer:

A.M. = 35

or x + 80 = 100 or X = 100 – 80 = 20

Hence, figure is = 20

Calculation of Median:

Median=Value of the item= 50th items

Value of 50th item lies in 20 – 30 class interval

= 20 + 7.41 = 27.41

Question 5.

The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Workers | Daily Income (Rs.) |

A | 120 |

B | 150 |

C | 180 |

D | 200 |

E | 250 |

F | 300 |

G | 220 |

H | 350 |

I | 370 |

J | 260 |

Answer:

Question 6.

Following information pertains to the daily income of 150 families Calculate the arithmetic means.

Income is (Rs) | No. of Families |

More than 75 | 150 |

More than 85 | 140 |

More than 95 | 115 |

More than 105 | 95 |

More than 115 | 70 |

More than 125 | 60 |

More than 135 | 40 |

More than 145 | 25 |

Answer:

AM = 120

= 120 – 3.7 = 116.3

Question 7.

The size of land holdings of 380 families in a village is given below. Find the median size of land holdings.

Size of Land Holdings (in acres) | Number of families |

less than 100 | 40 |

100 – 200 | 89 |

200 – 300 | 148 |

300 – 400 | 64 |

400 and above | 39 |

Answer:

Calculation of median size of land holding.

Median = Value of \(\frac{\mathrm{N}}{2}\)th item = \(\frac{380}{2}\)

= Value of 190th item

Value of 190th item lies in Class interval of 200 – 300.

Hence, Median

= 200 + 4.21

= 241.21 acres

Question 8.

The following series relates to the daily income of workers employed in a firm. Compute the

- Highest income of lowest 50% workers,
- Minimum income earned by the top 25% and
- Maximum income earned by loest 25% workers.

Daily Income (in Rs.) | Number of workers |

10- 14 | 5 |

15 – 19 | 10 |

20-24 | 15 |

25-29 | 20 |

30-34 | 10 |

35 – 39 | 5 |

Answer:

Median = Value of \(\frac{\mathrm{N}}{2}\)th item = \(\frac{65}{2}\)

= \(\frac{65}{2}\) = Value of 32.5 item

Value of 32.5 item lies in 24.5 – 29.5 class interval

= 24.5 + \(\frac{2.5}{4}\)

= 24.5 + 0.56

= 25.11 Ans.

b) Q_{1} = Value of \(\left(\frac{\mathrm{N}}{4}\right)\) th item

\(\frac{65}{2}\) = Value of 16.25 item

Value of 16.25 item lies in 19.5 – 24.5 class interval

= 19.5 + 0.42

= 19.92 Ans.

c) Q_{3} = Value of \(\frac{3 \mathrm{N}}{4}\) th item.

\(=\frac{65 \times 3}{4}=\frac{195}{4}\)

= Value of 48.75 item

Value of 47.75 lies in 24.5 – 29.5 class interval.

= 24.5 + 4.69

= 29.19 Ans.

Question 9.

The following table gives production yield in kg. per hectare of wheat of 150 farms in a village. Calculate the mean, median and mode production yield.

Answer:

Calculation of Mean, Median and Mode

Mean A. M. = 63.5

C = 3

= 63.5 + 0.32

Median = 63.82 Ans.

Medain = Value of \(\frac{N}{2}\) th value

Value of 75th item which lies in 62 – 65 class interval.

Hence the modal class is 62 – 65

= 62 + 1.67 – 63.37 Ans.

Model Class = 62 – 65 because it has the maximum frequency i.e., 36.

= 62 + \(\frac{6}{72-58}\) × 3

= 62 + \(\frac{6}{14}\) × 3

= 62 + 1.29 = 62.29

### 1st PUC Economics Measures of Central Tendency Additional Questions and Answers

**1st PUC Economics Measures of Central Tendency Very Short Answer Type Questions**

Question 1.

What do you mean by ‘Average’?

Answer:

Average value or central tendency is the critical value which represents all the items in a series.

Question 2.

Pocket expenditures of 6 students are (Rs) 6, 12, 18, 24, 30 and 36 respectively. Find out arithmetic mean?

Answer:

Question 3.

Name various kind, of stastistical averages.

Answer:

i) Mathematical average:

- Arithmetic mean
- Geometric mean
- Harmonic means

ii) Positional Averages:

- Median
- Partition value
- Mode

Question 4.

Give formula of weighted averages

Answer:

Question 5.

Give formula of calculating arithmetic averages of a continuous series?

Answer:

Question 6.

What is arithmetic mean?

Answer:

Arithmetic mean is a simple average of all items in a series it is simply mean value which is obtained by adding the values of all the items and dividing the total by the number of etc.

Question 7.

Define Median?

Answer:

Median is a centrally located value of series such that half of the values of the serives are above it and other half below it.

Question 8.

Define mode?

Answer:

Mode is the value which occurs mode frequently in the series.

Question 9.

Give formula for finding out median.

Answer:

M = size of \(\left[\frac{N+1}{2}\right]\) the item

M = Median N = No.of items

Question 10.

Give formula for estimating mode in frequency distribution series?

Answer:

1. First we find out the model class

2.

Z = Value of the mode

I_{1} = Lower limit of the model classes

i = Size of the model group.

Question 11.

Give formula for finding out median of a continuous series?

Answer:

I_{1} = Lower limit of the median class

C.f = Cumulative frequency of the classes

f = Preceding the median classes

i = size of the median class interval

Question 12.

Define Quartile?

Answer:

If a statistical series is divided into four equal parts then the end value of each part is called a quartile.

Question 13.

What does central Tendency refer to?

Answer:

Central Tendency refers to a central value or a representive value of statistical date. Measures of central tendency- arithmetic mean.

Question 14.

Enumerate any three essential, of a good average?

Answer:

- It should be simple
- It must be certain on character.
- It should be capable of further mathematical or algebraic treatment.

Question 15.

Name the three statistical measures of central Tendency?

Answer:

Three statistical measure of central tendency are:

- Arithmetic mean
- Median and
- Mode

Question 16.

Write down the formuala for calculating mean for ungrouped data?

Answer:

\(\bar{x}=\frac{\sum x}{N}\)

\(\overline{\mathrm{x}}\) Stands for arithmetic mean

Σ x stands for summation of all the observations in a series

N = Stands for number of observations

Question 17.

Write down an interesting property of AM

Answer:

The sums of deviations of items about arithmetic mean is always equal to zero symbolically.

Question 18.

Write down the formula for calculating weighted average (mean)?

Answer:

\(\overline{\mathrm{X}}_{\mathrm{w}}=\frac{\sum_{\mathrm{x}} \mathrm{W}_{\mathrm{x}}}{\sum \mathrm{W}}\)

\(\overline{\mathbf{X}}_{\mathrm{w}}\) = Weighted average (mean)

W = Weights

X = variables

Question 19.

Give a formula for calculating arithmetic mean of continuous series by step deviation method?

Answer:

\(\bar{x}=A+\frac{f d^{\prime}}{N} \times C\)

Question 20.

Give the formula for calculating corrected Arithmetic mean?

Answer:

Corrected arithmetic mean

Σ x (Wrong) + (correct value)

= \(-\left(\frac{\text { Uncorrected value }}{\mathrm{N}}\right)\)

Question 21.

What is the arithmetic mean of 5, 6, 7, 8 and 9?

Answer:

Arithmetic mean of 5,6,7, 8 and 9 is \(\frac{35}{5}\) = 7

Question 22.

When can the direct method of computation of arithmetic mean be used?

Answer:

The direct method of computation of arithmetic mean can be used when the items in the serial are less.

Question 23.

Write down the formula of calculating the weighted arithmetic mean of three related groups?

Answer:

\(\bar{X}_{1,2,3}=\frac{N_{1} \bar{X}+N_{2} \bar{X}_{2}+N_{3} \bar{X}_{3}}{N_{1}+N_{2}+N_{3}}\)

Question 24.

When will the weighted mean be greater than the simple arithmetic mean?

Answer:

The weighted arithmetic mean will be greater than the simple arithmetic mean when items of small values are given less weights and items of big values are given more weights.

Question 25.

Name any two positional values?

Answer:

- Median
- Mode.

Question 26.

What is median?

Answer:

Median is that positional value of the variable which divides the distribution into two equal parts. One part comprises all values greater than or equal to median values and the other comprises all values less than or equal to it.

Question 27.

What is the first and most important rule for calculating the median?

Answer:

The first and most important rule for calculating median is that data should be arranged in ascending or descending order.

Question 28.

Calculate the median of 5.7.6.6.1.8.10.12.4 and 3

Answer:

By arranging the date in ascending order we get 1.3.4.5.6.7.8.10.12

Median = value of \(\frac{N+1}{2}\) item = \(\frac{9+1}{2}\) = 5th item i.e. 6.

Question 29.

Calculate the median of 3.5.7 and 4

Answer:

Medain = size of \(\frac{N+1}{2}\) item

= \(\frac{4+1}{2}\) = 2.5 item = \(\frac{5+1}{2}\) = 6

Question 30.

Write down the formula for locating the median from median group?

Answer:

I_{1} = Lower limit of median group.

c. f. = Cumulative frequency of the class Preceding the median class.

f = Frequency of the median goup.

i = The class interval of the median group.

Question 31.

What is partition value?

Answer:

The value that divides the series into more than two parts is called partition value Q_{1} Q_{2} Q_{3} etc are partition value.

Question 32.

Write down the difference between averages and partition values.

Answer:

An average is representitive of shole series while quartiles are average of parts of series.

Question 33.

What are Percentiles?

Answer:

Percentiles are measures which divide the distribution to hundred equal parts there are 99 percentiles denoted by P_{1} P_{2}P_{3} P_{99}.

Question 34.

Suppose a person has secured 82 percentiles in a management entrance examination. What does it mean?

Answer:

It means that his position is below 18 percent of total candidates appeared in the examination.

Question 35.

What is mode?

Answer:

Mode is the most frequenctly observed date value it is denoted by M_{0}.

Question 36.

Calculate mode from the following discrete series?

Answer:

The value of mode is 30 because it is most frequently observed data.

Question 37.

What do you mean by unimodal data.

Answer:

Unimodal data is that which has only one model for example there is one modal in the following data.

2.23.4.4 and 5

It is unimodal because it has one mode 30.

**1st PUC Economics Measures of Central Tendency Short Answer Type Questions**

Question 1.

Write any four merits of mean?

Answer:

Merits mean are as follows:

1. Certainity:

Arithemetic mean is a certain value it has no scope for estimated values.

2. Based on all items:

Arithmetic mean is based on all items in a series it is therefore based on all the items in series it is therefore representative value of the different items.

3. Simplicity:

From the view point of calculation and the use of arithmetic mean is the simplest of all the measures of central tendency.

4. Stability:

Arithmetic mean is a stable measure of centrel tendency this is because charges in the sample of a series having minimum effect or the arithmetic average series.

Question 2.

Write any four Demerits of mean?

Answer:

Demerits of arithmetic average are as follows:

- Effect of extreme value – The main effect of it is that it gets distorted by extreme values

of the series. - Unsuitability – Arithmetic mean is a suitable measures in case of percentage or proportionate values.
- Misleading conclusion – Arithmetic mean sometimes offer misleading conclusions.
- Laughable conclusion – Arithmetic mean sometimes offer laughable conclusion.

Question 3.

Compare the arithmetic mean, median and mode as measure of central tendency. Describe situations where one is more suitable than the others?

Answer:

As compared to mean and median, mode is less suitable. Mean is simple to calculate its value is definite. It can be given algebraic treatment and is not affected by fluctuations of sampling median and is even more simple to calculate but is affected by fluctuations and cannot be given algebraic treatment.

Mode is the most popular item of a series but it is not suitable for most elementary studies because it is not based on all the observations of the series and is unrepresentative.

In case of arithmetic mean it is the numberical magnitude of the deviations that balances In case of median it is the number of values greater than the median which balance against the number or values less than the median the median, is always between the arithmetic mean and the mode.

Question 4.

The arithmetic mean is described as the centre of gravity of the distribution of values of the variable Explain?

Answer:

This is rightly said that arithmetic mean is the center of gravity of the distribution of values of the variable because it is the number obtained by dividing the total values of different items by their number. We use this method in our everyday life. It is a popular and most widely used measure.

Question 5.

Find mean and median for aft four values of the series and answer the question given at the end of table?

1. is median affected by extreme values?

2. is median a better method than mean

Answer:

1. No, median is not affected by extreme values

2. Yes, median is a better method then mean.

Question 6.

Prove that the sum of the squares of the and deviation from arithmetic mean is the least ie. less other squares of the deviation of observation taken from any values.

Answer:

For providing the statement given in the questions we take the following observations

We take another value and we observe that the same of squares of deviations taken from mean (3) less that taken from other value (4) less.

Question 7.

Prove with an example that the weighted arithmetic mean will be greater than the simple arithmetic mean when items of small values are given less weights and items of big values are given more weights?

Answer:

In order to prove the statements given in the example we take the following table.

Question 8.

With the help of following data prove that the sum of the deviations of items about arithmetic mean is always equal to zero.

X : 4, 6, 8, 10 and 12

Answer:

From the given data it has been proved that the sum of the deviation of item, arithmetic mean is equal to zero.

Symbolically

Σ cx – \(\frac{2}{x}\) = 0

Question 9.

Calculate median from the following series.

Answer:

Median = Size of \(\frac{N+1}{2}\) th item

= \(\frac{30+1}{2}\) = value of 15.5 th item

Value of 15.5 the item = 30

Hence median is equal to 30.

Question 10.

Write down the merits of median?

Answer:

Merits of median:

Following are the merits of median.

- Median is definite
- It is easy to calculate and understand.
- It can also be determined graphically
- It is not affected by extreme values.
- It can be calculated in the absence of one of the item.
- It is helpful in qualitative facts such as ability stability etc.

Question 11.

Write down the demerits of median?

Answer:

Demerits of median:

Following are the demerits of median:

- In it all the items of a series are not given equal importance.
- If the number of items are given the correct value of median cannot be calculated.
- It is affected by fluctuation of sampling.
- It cannot be give further algebraic treatment.
- Data needs to be arranged in ascending of descending order.

Question 12.

Why is arithmetic mean the most popular measure of the central tendency?

Answer:

Arithmetic mean is the most popular measure of the central tendency due to following reasons:

- It is easy to calculate and understand arithmetic mean.
- Arithmetic mean is based on all the values of the serial.
- Arithmetic mean is stable measure or central tendency it is because change in the sample of a series have minimum effect on the arithmetic averages.

Question 13.

Daily income of eight families are given below. Calculate average daily family income.

Answer:

Sl.No. | Daily income (Rs) X |

1 | 170 |

2 | 500 |

3 | 250 |

4 | 700 |

5 | 300 |

6 | 400 |

7 | 200 |

8 | 350 |

N = 8 | Σ x = 2870 |

\(\overline{\mathrm{X}}=\frac{\sum \mathrm{x}}{\mathrm{N}}=\frac{2870}{8}\)

= Rs. 358.75

Question 14.

The average height of 50 student’s in a class are 61 inches and that of 70 students of another class is 58 inches Calculate combined average height of both the classes?

Answer:

Given N_{1} = 50, \(\overline{\mathbf{X}}_{1}\) = 61

N_{2} = 70, \(\overline{\mathbf{X}}_{2}\) = 58

= 59.25 inches

Question 15.

The following table gives the daily income of ten workers in a factory. Find the arithmetic mean.

Workers | Daily Income (Rs) |

A | 120 |

B | 150 |

C | 180 |

D | 200 |

E | 250 |

F | 300 |

G | 220 |

H | 350 |

I | 370 |

J | 260 |

Answer:

\(\overline{\mathbf{X}}\) = X_{1} + X_{2} + X_{3} + X_{4}…….+ X_{N}

\(=\frac{120+150+180+200+250+300+220+350+370+260}{10}\)

\(=\frac{2400}{10}\) = Rs. 240

Question 16.

Calculate the median from the following data.

Marks obtained | No. of Students |

20 | 5 |

40 | 10 |

60 | 20 |

80 | 15 |

100 | 10 |

Answer:

Median = Value of \(\frac{\mathrm{N}+1}{2}\) th item

= Value of \(\frac{\mathrm{60}+1}{2}\) th item

i.e value of 30.5th item

Value of 30.5 is 60

Hence Median = 6.

Question 17.

Monthly incomes of five families are given below. Calculate mean by shortcut method. 6550, 7550, 9550, 4550 and 8000

Answer:

Assumed mean = 5550

\(\overline{\mathbf{X}}\) = A.M + \(\frac{\sum \mathrm{d}}{\mathrm{N}}\)

= 5550 + \(\frac{8450}{5}\)

= 5550 + 1690

= 7240.

**1st PUC Economics Measures of Central Tendency Long Answer Type Questions**

Question 1.

The marks obtained by 30 students of PUC first year of a college in economics paper are given in the college in economics paper are given in the following table. Find the mean marks obtained by the students.

Answer:

Direct Method:

= 59.3 marks

Question 2.

The table below gives the percentage distribution of female teachers in the primary schools of rural area of Karnataka State. Find mean percentage of female teachers by assumed mean method and step deviation Method.

Answer:

Direct Method:

= 39.71

Step deviation method:

= 39.71 the mean percentage of female teachers is 39.71

Question 3.

Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of workers.

Answer:

Question 4.

The following table gives the literarcy rate (in %) of 35 cities. Find the mean literarcy rate. Literacy rate in

Answer:

Question 5.

The table below shows the daily expenditure on food of 25 households in a locality.

Answer:

The mean daily expenditure on food is Rs. 211 per Household

Question 6.

The following series relates to the daily income of workers employed in a firm. Compute the highest income of lowest 50% of workers, ie; median.

Answer:

Since the series of data given is of inclusive type, first we shall convert the class intervals into exclusive type using a correction factor.

C.F – LL of a CI – UL of the previous CI

= \(\frac{15-14}{2}\) = 0.5

Where LL = Lower Limit

UL = Upper Limit

CI = Class Interval

CF = Cumulative frequency

Deduct 0.5 from lower limit and add 0.5 to the upper limits

Median class = Size of \(\left[\frac{\mathrm{N}}{2}\right]\)^{th} item

= \(\left[\frac{\mathrm{65}}{2}\right]\)^{th} item = 32.5 th item

so, it lies between 24.5 = 29.5

Medain = \(\mathrm{L}+\left[\frac{\frac{\mathrm{N}}{2}-\mathrm{Cf}}{\mathrm{f}}\right] \times \mathrm{i}\)

Where L – Lower limit

C.f = Cumulative frequency of class preceding the median class

L – Length of the median class

N = Total of frequency

f = frequency of the median class

Median class = 24.5 – 29.5

I = 24.5; i = 5; c.f = 30; N = 65; f = 20

Medain = 24.5 + \(\left[\frac{\frac{65}{2}-30}{20}\right]\) × 5

Medain = Rs. 25.125

Highest income of lowest 50% of workers