Tamilnadu State Board New Syllabus Samacheer Kalvi 12th Business Maths Guide Pdf Chapter 9 Applied Statistics Ex 9.1 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 12th Business Maths Solutions Chapter 9 Applied Statistics Ex 9.1

Question 1.

Define Time series.

Solution:

When quantitative data are arranged in the order of their occurrence, the resulting series is called the Time Series.

Question 2.

What is the need for studying time series?

Solution:

Time series analysis is one of the statistical methods used to determine the patterns in data collected for a period of time. Generally, each of us should know about the past data to observe and understand the changes that have taken place in the past and current time. One can also identify the regular or irregular occurrence of any specific feature over a time period in a time series data.

Question 3.

State the uses of time series.

Solution:

Question 4.

Mention the components of the time series.

Solution:

There are four types of components in a time series. They are as follows

(i) Secular Trend

(ii) Seasonal variations

(iii) Cyclic variations

(iv) Irregular variations

Question 5.

Define secular trend.

Solution:

Secular Trend: It is a general tendency, of time series to increase or decrease, or stagnates during a long period of time. An upward tendency is usually observed in the population of a country, production, sales, prices in industries, the income of individuals, etc., A downward tendency is observed in deaths, epidemics, prices of electronic gadgets, water sources, mortality rate, etc…

Question 6.

Write a brief note on seasonal variations

Solution:

As the name suggests, tendency movements are due to nature which repeats themselves periodically in every season. These variations repeat themselves in less than one year time. It is measured in an interval of time. Seasonal variations may be influenced by natural force, social customs and traditions. These variations are the results of such factors which uniformly and regularly rise and fall in the magnitude. For example, selling of umbrellas’ and raincoat in the rainy season, sales of cool drinks in the summer season, crackers in the Deepawali season, purchase of dresses in a festival season, sugarcane in Pongal season.

Question 7.

Explain cyclic variations.

Solution:

Cyclic Variations: These variations are not necessarily uniformly periodic in nature. That is, they may or may not follow exactly similar patterns after equal intervals of time. Generally, one cyclic period ranges from 7 to 9 years and there is no hard and fast rule in the fixation of years for a cyclic period. For example, every business cycle has a Start- Boom-Depression- Recover, maintenance during booms and depressions, changes in government monetary policies, changes in interest rates.

Question 8.

Explain the stratified random sampling with a suitable example.

Solution:

These variations do not have a particular pattern and there is no regular period of time of their occurrences. These are accidental changes which are purely random or unpredictable. Normally they are short-term variations, but their occurrence sometimes has its effect so intense that they may give rise to new cyclic or other movements of variations. For example floods, wars, earthquakes, Tsunami, strikes, lockouts, etc.

Question 9.

Define seasonal index.

Solution:

Seasonal Index for every season (i.e) months, quarters, or year is given by

Seasonal Index (S.I) = \(\frac{\text { Seasonal Average }}{\text { Grand average }}\) × 100

Where seasonal average is calculated for month, (or) quarter depending on the problem and Grand Average (G) is the average of averages.

Question 10.

Explain the method of fitting a straight line.

Solution:

(i) The straight-line trend is represented by the equation Y = a + bX …………. (1)

Where Y is the actual value, X is time, a, b are constants.

(ii) The constants ‘a and ‘b’ are estimated by solving the following two normal Equations

ΣY = n a + b ΣX ………… (2)

ΣXY = a ΣX + b ΣX² …………… (3)

Where ‘n’ = a number of years given in the data.

(iii) By taking the mid-point of the time as the ori-gin, we get ΣX = 0

(iv) When ΣX = 0, the two normal equations reduces to

ΣY = na + b(0) ; a = \(\frac { ΣY }{n}\) = \(\bar { Y}\)

ΣXY = a(0) + bΣX² ; b = \(\frac { ΣXY }{ΣX^2}\)

The constant ‘a’ gives the mean of Y and ‘b gives the rate of change (slope),

(v) By substituting the values of ‘a and ‘b’ in the trend equation (1), we get the Line of Best Fit.

Question 11.

State the two normal equations used in fitting a straight line.

Solution:

The normal equations used in fitting a straight line are

ΣY = na + b ΣX and ΣXY = a ΣX + b ΣX^{2}

Where n = number of years given in the data,

X = time

Y = actual value

a, b = constants

Question 12.

State the different methods of measuring trends.

Solution:

Following are the methods by which we can measure the trend.

(i) Freehand or Graphic Method

(ii) Method of Semi-Averages

(iii) Method of Moving Averages

Question 13.

Compute the average seasonal movement for the following series

Solution:

Computation of seasonal. Index by the method of simple averages.

Question 14.

The following figures relates to the profits of a commercial concern for 8 years

Find the trend of profits by the method of three yearly moving averages.

Solution:

Question 15.

Find the trend of production by the method of a five-yearly period of moving average for the following data:

Solution:

I Computation of five – yearly moving averages

Question 16.

The following table gives the number of small- scale units registered with the Directorate of Industries between 1985 and 1991. Show the growth on a trend line by the free hand method.

Solution:

Question 17.

The annual production of a commodity is given as follows:

Fit a straight line trend by the method of least squares

Solution:

Therefore the required equation of the straight line

Y = a + bx

Y = 169.428 + 3.285 X

⇒ Y = 169.428 + 3.285 (x – 1998)

Question 18.

Determine the equation of a straight line which best fits the following data

Compute the trend values for all years from 2000 to 2004

Solution:

Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by

Y = a + bX

Y = 54 + 5.4X

Y = 54 + 5.4 (x – 2002)

The trend value can be obtained by when x = 2000

Yt = 54 + 5.4 (2000 – 2002)

Y = 54+ 5.4 (-2)

= 54 – 10.8

= 43.2

When x = 2001

Yt = 54 + 5.4 (2001 – 2002)

Y = 54 + 5.4 (-1)

= 54 – 5.4

= 48.6

When x = 2002

Yt = 54 + 5.4 (2002 – 2002)

V = 54 + 5.4 (0)

= 54

When x = 2003

Yt = 54 + 5.4 (2003 – 2002)

Y = 54 + 5.4 (1)

= 54 + 5.4

= 59.4

When x = 2004

Yt = 54 + 5.4 (2004 – 2002)

Y = 54 + 5.4(2)

= 54 + 10.8

= 64.8

Question 19.

The sales of a commodity in tones varied from January 2010 to December 2010 as follows: in year 2010 Sales (in tones)

Fit a trend line by the method of semi-average.

Solution:

Since the number of years is even (twelve). We can equally divide the given data it two equal parts and obtain averages of the first six months and the last six.

Question 20.

Use the method of monthly averages to find the monthly indices for the following data of production of a commodity for the years 2002, 2003, and 2004.

Solution:

Question 21.

Use the method of monthly averages to find T the monthly indices for the following data of production of a commodity for the years 2002, 2003, and 2004.

Solution:

Computation of Seasonal Index by the method of simple averages

Question 22.

The following table shows the number of salesmen working for a certain concern:

Use the method of least squares to fit a straight line and estimate the number of salesmen in 1997

Solution:

Computation of trend values by the method of least squares (ODD Years)

Therefore, the required equation of the straight line trend is given by

Y = a + bX

Y = 48.8 + 2X

(i.e) Y = 48.8 + 2 (x – 1994)

When x = 1992

Yt = 48.8 + 2 (1992 – 1994)

Y = 48.8 + 2 (-2)

= 48.8 – 4

= 44.8

When x = 1993

Yt = 48.8 + 2 (1993 – 1994)

Y = 48.8 + 2 (-1)

= 48.8 – 2

= 46.8

When x = 1994

Yt = 48.8 + 2 (1994 – 1994)

Y = 48.8 + 2(0)

= 48.8

When x = 1995

Yt = 48.8 + 2 (1995 – 1994)

Y = 48.8 + 2(1)

= 50.8

When x = 1996

Yt = 48.8 + 2 (1996 – 1994)

Y = 48.8 + 2 (2)

= 48.8 + 4

= 52.8

When x = 1997

Yt = 48.8 + 2 (1997 – 1994)

Y = 48.8 + 2 (3)

= 48.8 + 6

= 54.8