CET

1. GUIDANCE &
COUNSELLING
Ms. K. Lavanya
MSc(N)-CHN
Associate Professor

2. Introduction:
• Guidance & counselling are twin concepts & have emerged as essential elements of
every educational activity.
• Guidance & counselling are not synonymous term. Counselling is a part of
guidance.
• Guidance, in educational context, means to indicate, point out, show the way, lead
out & direct.
• Counselling is a specialized service of guidance. It is the process of helping
individuals learn more about themselves & their present & possible future situations
to make a substantial contribution to the society.

3. Guidance:

4. Definition:
• Guidance is a process of dynamic interpersonal relationships designed to
influence the attitudes and subsequent behavior of a person. –Good
• Guidance is assistance made available by personally qualified and adequately
trained personnel to an individual of any age to help him manage his own life
activities, to develop his own points of view, make his own decisions and
carry his own burdens- Crow and Crow.
• Guidance as a process through which an individual is able to solve their
problems and pursue a path suited to their abilities and aspirations – JM
Brewer.

5. • Guidance is a continuous process of helping the individual development in
the maximum of their capacity in the direction most beneficial to himself and
to society- stoops and wahlquist.

6. Purpose of Guidance
1. To bring confidence in selecting appropriate course of action for adjustment in
various walks of life.
2. Helping in a balanced development.
3. To help to determine the courses most appropriate to their needs and abilities.
4. To plan the future in the individual’s line of interest, abilities and social needs.

8. Purpose of guidance and counselling
• The purpose of Guidance and Counselling can be related to 1) Individual and 2) Society.
1) Individual Related Purpose
(i) To help students recognize and develop their potentialities to achieve their educational aims and
objectives and improve academically and to develop positive attitude.
(ii) To help students aware about vocational and career opportunities available regionally, nationally
and internationally so as to help them make informed decisions from among various choices.
(iii) To help students for all round personal and social development on the basis of their interests,
abilities and resources.
(iv) To help students recognize their capacities, develop self confidence and adjust to academic,
school, family and personal problems.

9. 2) Society Related Purpose
(i) To develop good citizenship in students
(ii) To develop positive attitude towards family life and the society
(iii) To help in proper and best utilization of resources.

10. Other functions of guidance and counselling…
• To provide optimum development & well-being for individual.
• To help individuals adjust to themselves & the society.
• To help people understand themselves in relation to the world.
• To aid individuals in efficient decision making.
• To help individuals plan for a productive life in their social context by focusing on their
assets, skills, strengths & possibilities for further development.
• To bring about changes in the attitude & behavior of individuals.

11. Characteristics of Guidance
It is a process as
It helps every individual to help himself to recognize and use his inner resources,
 To set goals
 To make plans
 To work out his own problems of development
It is a continuous process Choice & problem points are the distinctive concerns of guidance
It is the assistance to the individual in the process of development rather than a direction of that
development
Guidance is a service meant for all Guidance is both generalized & a specialized service.

13. Short cut or Assumed mean method:
• When observations in data set are large in size, it is a laborious work to find
mean. To avoid this difficulty, short cut method is adopted.
• Assume arbitrary mean i.e., an value from data set (which will simplify the
calculations) and subtract this assumed mean from each observation.
• We get what is known as differences or deviations.
• Obtain mean for deviations by usual method.

14. Contd….
• Observations
• Original data: X1, X2, ………Xn
• Differences or X1-a, X2-a, …….. Xn-a
• Deviations: d1, d2,…..dn
• Where a is any value from dataset.
• Mean for deviations(d) = sum d/n. Thus, Mean of original data(X)=a+d

15. Example:
• In a series of 10 postmorterms following observations regarding weight (in
gms) of liver were found.
• 1420 1405 1425 1410 1415
1435 1430 1415 1445 1430

16. Solution:
• Let us consider a=1420
• d= x-1420: 0 -15 5 -10 -5 15 10 -5 25 10
• 𝑑 =
𝑠𝑢𝑚𝑑
𝑛
=30/10 =3
• Thus x= a+d =1420+3=1423

17. Computation of grouped data
• In Statistics, data plays a vital role in estimating the different types of parameters. To
draw any conclusions from the given data, first, we need to arrange the data in such a
way that one can perform suitable statistical experiments. We know that data can be
grouped into two ways, namely, Discrete and Continuous frequency distribution.

18. Discrete frequency distribution:
• Suppose we have X1, X2, …….. Xn observations with corresponding
frequencies f1, f2,…..,fn. The AM is defined as
• 𝑥 =
𝑓1𝑥1+𝑓2𝑥2+⋯+𝑓𝑛𝑥𝑛
𝑓1+𝑓2
+…+𝑓𝑛
• In notation form, we have
• MeanX= ∑(f.x)/ ∑f
= ∑(f.x)/N
= Sum (Frequency×observation)
• Total Frequency

19. Calculate the average number of children per
family from the following data:
NO: of children No: of families
0 30
1 52
2 60
3 65
4 18
5 10
6 05

20. Solution:
NO: of Children
(X)
NO: of families
(f)
Total NO: of Children
(f.x)
0 30 0×30=0
1 52 1×52=52
2 60 2×60=120
3 65 3×65=195
4 18 4×18=72
5 10 5×10=50
6 5 6×5=30
Total 240 519

21. Contd….
• X=
• MeanX= ∑(f.x)/ ∑f
= 519/240=2.165

22. Continuous frequency distribution:
• In continuous frequency distribution, the frequency is not associated with
any specified single value but spread over entire class.
• It creates difficulty for finding mid values X1, X2,….,Xn. To overcome this
difficulty, we make a reasonable assumption that the frequency is associated
with mid-value of class, or the frequency is distributed uniformly over the
entire class.
• Mean (X) = Sum(f.x)/ Sum(f)

23. The following are different steps to calculate average
for continuous frequency distribution
• Step 1- Write all class intervals serially in the first column and corresponding
frequency in the second column.
• Step 2- The mid values of each class interval are obtained by adding lower
and upper class interval and dividing resultant quantity by 2 and put these
values in third column.
• Step 3- Multiply each ‘f’ by corresponding X and write this product in fourth
column. The addition of this column gives sum(fx). i.e ∑f.x.

24. Notation form:
• X= Sum of fourth column
Sum of second column
= Sum (f.x)
Sum(f)

25. Example:
• Find the average age (in years) at the time of death in city A.
Age Interval NO: of Deaths
0-10 16
10-20 09
20-30 20
30-40 11
40-50 07
50-60 12
60-70 09
70-80 04
80-90 02

26. Solution:
Age Interval No: of Deaths
(f)
Mid Values
(x)
Frequency Observation
(f.x)
0-10 16 05 16×5=80
11-20 09 15 9×15=135
21-30 20 25 20×25=500
31-40 11 35 11×35=385
41-50 07 45 7×45=315
51-60 12 55 12×55=660
61-70 09 65 9×65=585
71-80 04 75 4×75=300
81-90 02 85 2×85=170
TOTAL 90 3130

27. • X= sum(fx)
sum(f)
= 3130/90
=34.78
Average age at death is 34.78 years

28. 2. MEDIAN
• The mean is unduly affected by extreme observations and cannot be
calculated for distribution with open end class and qualitative variables like
honesty, sex, religion etc.
• To overcome these drawbacks, we use other measures of central tendency
like median.

29. Definition:
• When all the observations of a variable are arranged in either ascending or
descending order, the middle observation is known as median. It divides the
whole data into two equal portions.
• In other words, 50% of the observations will be smaller than the median
while 50% of the observations will be larger than it.

30. Computation of Median:
Ungrouped Data:
• As discussed above, the median is one of the measures of central tendency,
which gives the middle value of the given data set.
• While finding the median of the ungrouped data, first arrange the given data
in ascending order, and then find the median value.

31. • If the total number of observations (n) is odd, then the median is (n+1)/2 th
observation.
• If the total number of observations (n) is even, then the median will be average of
n/2th and the (n/2)+1 th observation.

32. Example:
For example, 6, 4, 7, 3 and 2 is the given data set.
• To find the median of the given dataset, arrange it in ascending order.
• Therefore, the dataset is 2, 3, 4, 6 and 7.
• In this case, the number of observations is odd. (i.e) n= 5
• Hence, median = (n+1)/2 th observation.
• Median = (5+1)/2 = 6/2 = 3rd observation.
• Therefore, the median of the given dataset is 4

33. Calculation for grouped data
• In a grouped data, it is not possible to find the median for the given observation by
looking at the cumulative frequencies. The middle value of the given data will be in
some class interval. So, it is necessary to find the value inside the class interval that
divides the whole distribution into two halves.
• we have to find the median class.
• To find the median class, we have to find the cumulative frequencies of all the classes
and n/2. After that, locate the class whose cumulative frequency is greater than (nearest
to) n/2. The class is called the median class.

34. Formula:

35. Example:

36. Solution:
• To find the median height, first, we need to find the class intervals and their corresponding frequencies.
• The given distribution is in the form of being less than type,145, 150 …and 165 gives the upper limit. Thus,
the class should be below 140, 140-145, 145-150, 150-155, 155-160 and 160-165.
• From the given distribution, it is observed that,
• 4 girls are below 140. Therefore, the frequency of class intervals below 140 is 4.
• 11 girls are there with heights less than 145, and 4 girls with height less than 140
• Hence, the frequency distribution for the class interval 140-145 = 11-4 = 7
• Likewise, the frequency of 145 -150= 29 – 11 = 18
• Frequency of 150-155 = 40-29 = 11
• Frequency of 155 – 160 = 46-40 = 6
• Frequency of 160-165 = 51-46 = 5

37. Therefore, the frequency distribution table along
with the cumulative frequencies are given below:
Class Intervals Frequency Cumulative Frequency
Below 140 4 4
140 – 145 7 11
145 – 150 18 29
150 – 155 11 40
155 – 160 6 46
160 – 165 5 51

38. Contd….
• Here, n= 51.
• Therefore, n/2 = 51/2 = 25.5
• Thus, the observations lie between the class interval 145-150, which is called the
median class.
• Therefore,
• Lower class limit = 145
• Class size, h = 5
• Frequency of the median class, f = 18
• Cumulative frequency of the class preceding the median class, cf = 11.

39. • Now, substituting the values in the formula, we get
• Median=145+(25.5−1118)×5
• Median = 145 + (72.5/18)
• Median = 145 + 4.03
• Median = 149.03.
• Therefore, the median height for the given data is 149. 03 cm.

40. Practice Problem

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