CHSE Odisha Class 12 Foundations of Education Unit 4 Educational Statistics Questions and Answers

Odisha State Board CHSE Odisha Class 12 Foundations of Education Solutions Unit 4 Educational Statistics Questions and Answers.

CHSE Odisha 12th Class Foundations of Education Unit 4 Educational Statistics Questions and Answers

Objective Type Question and Answers

Question 1.
What is the formula to find out the mean?
(a) \(\frac{\sum F x}{N}\)
(b) \(\frac{F x}{N}\)
(c) L+\(\frac{N_2 F}{M}\)
(d) H.S.-L.S.
Answer:
(a) \(\frac{\sum F x}{N}\)

Question 2.
Estimate the mode of the scores:
8, 9, 8, 9,10,11,8,9,10
(a) 9
(b) 10
(c) 11
(d) 8
Answer:
(d) 8

Question 3.
What is a formula to find median?
(a) H.S. – L.S.
(b) \(\frac{Q_3-Q_1}{Q}\)
(c) 3 median – 2 mean
(d) None of the above
Answer:
(c) 3 median – 2 mean

Question 4.
4. Estimate the mean of the scores: 8,9,10,11,12,13
(a) 8.5
(b) 10.5
(c) 11.5
(d) 12.5
Answer:
(b) 10.5

Question 5.
5. Estimate the class interval of25-29.
(a) 4
(b) 3
(c) 5
(d) None
Answer:
(c) 5

Question 6.
6. Estimate the mean of the scores:
2, 8, 10, 25,45
(a) 10
(b) 20
(c) 30
(d) 40
Answer:
(a) 10

Question 7.
7. Select the formula to find out the median?
(a) A.M.+\(\frac{\sum F d}{N}\) Xi
(b) U\(\frac{\frac{N}{2}-F}{f m}\) xi
(c) 3 median – 2 mean
(d) None
Answer:
(b) U\(\frac{\frac{N}{2}-F}{f m}\) xi

Answer in single word / single sentence

Question 1.
How can you find out Mode?
Answer:
The mode can be calculated as 3 median – 2 mean.

Question 2.
What is Mean?
Answer:
The mean is the sum of the group of scores divided by the number of scores.
M = \(\frac{\sum x}{N}\)

Question 3.
What do you mean by Median?
Answer:
The median is the midpoint or mid-value of the distribution of scores.

Question 4.
Give one use of Mean?
Answer:
Mean is used when a quick and easily computerized measure of central tendency is needed.

Question 5.
Give one use of Mode?
Answer:
Mode is used when the quickest of the central tendency is needed.

Question 6.
What is the limitation of the Median?
Answer:
A median is an algebraic measure and hence not suitable for the function of algebraic treatment.

Question 7.
How can you calculate the Mean?
Answer:
mean = L +\(\left|\frac{\frac{N}{2}-F}{F m}\right|\) Xi

Question 8.
What is the limitation of Mode?
Answer:
Mode is not based on all the observations in a series.

Question 9.
What is the formula for finding out the mean of the ungrouped data?
Answer:
Mean = \(\frac{\sum X}{N}\)

Question 10.
What is the mode of an ungrouped set score?
Answer:
The score which occurs frequently or number of times is known as a mode.

Question 11.
How can you measure the median in a simple method
Answer:
Median = \(\left(\frac{N+1}{2}\right)^{t h}\) Term

Question 12.
Find out the median of the following scores:
5.4, 3, 7, 8, 10, 4,6
Answer:
Arrange is chronological order :
3, 4, 5, 6, 7, 8, 9, 10
Median = \(\left(\frac{N+1}{2}\right)^{t h}\) Score
= \(\frac{8+1}{2}\)
= \(9 / 2\)
= 4.5^th
= \(\frac{7+8}{2}\)
= \(\frac{15}{2}\)
=7.5

Question 13.
what is the midpoint of class interval: 1.3 -1.7?
Answer: The midpoint of C, I.
= 1.3- 1.7= 1.5

Question 14.
What is Mode? Give example.
Answer:
The mode is the score that occurs frequently in a group.
Example – 5, 6, 7, 8,4, 8, 9 Here mode is = 5.

Question 15.
Find out H.S.&L.S. of 14.5 -19.5
Answer:
14- 19 & 15-20

Question 16.
find out the mean: 40, 30,20, 80,60
Answer:
Mean = \(\frac{\sum X}{N}\)
= \(\frac{160}{5}\)
= 32

Question 17.
Find out H.S. & L.S. of the three class intervals 100 – 109,10 – 14, 80 – 84
Answer:

C.I	L.S.	H.S.
100 – 109	99.5	109.5
10-14	9.5	14.5
80-84	79.5	84.5

Very Short Type Questions With Answer

Question 1:
Explain how the knowledge of statistics is helpful to a teacher.
Answer:
1. Statistics helps the teacher to draw general conclusions.
2. It enables the teacher to predict the future performance of the pupils.

Question 2:
What is measures of central tendency?
Answer:
Methods of finding out the central values or average value of a statistical series of quantitative information mean, median and mode are the measures of central tendency, to find out the arithmetic averages.

Question 3:
Explain the different steps for preparing a frequency distribution table.
Answer:
The following are the steps of frequency distribution:

• Find out the range,
• Deciding the length of class interval.
• Making tallies to find out the frequency of scores with each class.
• Making the total number of tallies in each class.

Short Type Questions With Answers

Question 1:
1. What is statistics? What is measure of central tendency?
Answer:
Statistics is a branch of mathematics. It is a science dealing with numerical facts collected systematically with the purpose of action and study. So statistics is a science dealing with the collection, analysis and interpretation of data. It always deals with numerical facts and figures. The data are collected from various sources, then it is analyzed, and tabulated for interpretation. Now, statistics is used to measure population growth in education, agriculture and industry in finding out the numerical facts and figures statistics is used.

Measures of Central Tendency

Measures of central tendency is the method of finding out the central values or average value of statistical series of quantitative information mean, median and mode are the measures of central tendencies. Mean is the arithmetical average of the scores of a group. Median is the score point in a distribution below and above which half of the cases live. Mode means the score which occurs most frequently.

Question 2:
Write the function of statistics in education.
Answer:

• Statistics helps in the collection, presentation of data in a systematic manner.
• It helps in classification.
• It provides a technique for making comparison of examination results among the students.
• It helps to understand complex data by simplifying it.
• It helps to study the relationship between different results of examinations.

Question 3:
Explain the use of mean and Median.
Answer:
Use of Mean :

• When the distribution is symmetrical score and are uniformly distributed mean is the centre of gravity have the same value and other scores distribute it to its distribution
• Mean is the most stable measure of central tendencies and is often in demand at the greatest statistical calculations.
• The mean has the greatest stability so when a measure of central tendency with the greatest stability is needed the mean is used.

Use of Median :

The Median is used when the exact midpoint the 50 % of the distribution is desired.

• It is scored that would markedly affect the mean.
• When it is desired that certain scores should influence the central tendency, the median is used.
• It is used when the distribution has no upper or lower class interval of complicated length.

Long Type Questions With Answers

Question 1.
What is statistics? What is measures of E.T. ? Explain.
Answer:
Statistics is a branch of mathematics. It is a science dealing with numerical facts collected systematically with the purpose of action and study.

So statistics is a science dealing with the collection, analysis and presentation of data and interpretation. Statistics always deals with numerical facts and figures. The data are collected systematically from various sources then it is analysed, and tabulated for interpretation. How statistics is used to measure population, growth in education, agriculture and industry.

Helpful to a Teacher:

•  It helps the teacher to provide the most exact type of description.
• It makes the teaching definite and exact is procedures and thinking.
• It enables the teacher to draw general conclusions.
• It enables the teacher to predict to future performances of the pupils.

Functions of statistics:

• Statistics help into the collection and presentation the data in a systematic manner.
• It helps to understand complex data by simplifying it.
• It helps to classify data.
• It provides a technique for making comparisons of exams and results among students.
• It helps to study the relationship between different results of examination.

Needs :

• Statistical methods are used for standardisation of various tests and measures like – achievement tests in various subjects, like intelligence tests, interest, aptitude scales and various other measures of personality measurement.
• The scores obtained from various tests and measures are always relative. Statistical methods help their proper presentation, comparison, analysis and interpretation, and statistics help better.
• To make predictions regarding to further progress.
• Statistics is used for the ID of our students to study their interest and I.Q.
• To compare the variability some Mean, Median and Mode and its central tendency.

What is the Measures of C.T :
Central value or central tendency means the value which lies in the centre or middle of the distribution. Thus it is located somewhere in between two extreme values in the distribution. In simple words, it is the value of a variable around which other values are distributed
A measure of C.T. of the measure of a single typical value which is the best representative of the whole group.

Measures of CT is the methods finding at the central values on the average value of a statistical series of quantitative information. Mean median and mode are the measures of central tendencies.

Mean is the score point in a distribution in between and above which half of the cases like a sum total of the score divided by its number is mean.
So μ = \(\frac{\sum X}{N}\)

where ∑ = sum total
X = individual scores
N = total no. of frequencies

Mean = \(\frac{\text { the sum of all the values }}{\text { the number of values }}\)

Suppose, 5 boys of a class are selected for the cricket team and their ages are 15 16 16 17 17 18.
We can find out the arithmetic mean of the ages of students by adding the ages of 5 students and then dividing that sum by the number of boys 5.

Arithmetic mean \(\frac{75}{5}\) = 15

Calculation of mean by the short method:

Marks	No of students	Midpoints	Deviations	Fd
C l	F	(m)	(d)
0-10	6	5	-30	-180
10-20	14	15	-20	-280
30-40	16	25	-10	-160
40-50	27	35	o	0
50-60	22	45	10	220
	15	55	20	300
	N=100			F d = -100

Arithmetic mean = \(\overline{\mathbf{X}}\) = A = \(\frac{\sum \mathrm{fd}}{\mathbf{N}}\)
Where A = Arithmetic means 35
∑Fd = Total deviations = -100
N=∑f=100
\(\overline{\mathbf{X}}\)
= 35+\(\frac{-100}{100}\)
=35-1
=34

Median: Median is the mid value of the middle item of the series when the series are arranged in ascending or descending order of magnitude, smallest to longest or largest to smallest.

Median = value of \(\left(\frac{N+1}{2}\right)^{\text {th }}\) item

For example – if the ages of five boys are 7, 8, 9, 10, 11

Median age will be the age of \(\left(\frac{N+1}{2}\right)^{\text {th }}\)
Boy = \(\left(\frac{S+1}{2}\right)\) boy

i.e. the 3rd boy which is equal to 9 years.
Example: Find out the median of the grouped data.

C.T.	F	Count
45-50	1	30
41-50	2	29
35-40	2	27
31-34	2	25
26-30	4	23
21-25	5	19
16-20	1	14
11-15	6	13
6-10	4	7
1-05	3	3
	N=31

Here L= 25 \(\frac{N}{2}\)
= \(\frac{30}{2}\)
= 15

F= I4 Fm=5 i=5
Median = 20.5+\(\frac{15-14}{5}\) x 5
= 20.5+\(\frac{1}{5}\) x5
= 20.5+\(5 / 5\)
= 20.5 + 1
= 21.5

Mode = mode is the frequently occurring value in a distribution.
Example = Calculate the mode from the following data of the marks obtained by 20 students 15, 10,25, 30, 25,40
Mode = 25
Formula = 3 median – 2 mean = mode
(1) Use of Mean: Mean is used:

• When the distribution is systematic, scores are distributed. Mean is the centre of gravity
  of the scores of distribution.
• Mean is the most stable measure of C.T. and greater statistical calculation.
• When the measure of C.T. with its greatest stability is needed the mean is used.

(2)The use of Median :

• The Median is used when the exact midpoint of the distribution is desired.
• It is scores which would markable affect the mean.
• When it is desired that certain scores should be with CT median is used.
• It is used when the distribution has no upper or lower interval of complicated length. Use of Mode: Mode is used when the quickest of the central tendency is needed.

Question 2.
Find out mean and median of the distribution?

C.L	F
10 – 14	4
15 – 19	5
20 – 24	9
25 – 29	18
30 – 34	11
35 – 39	5
40 – 44	6
45 – 49	3
50 – 54	1
	N=60

Answer:

C.I.		F	d1	Fd1
15-19	7	-3	-21	-39
20-24	5	-2	-10
25-29	8	-1	-8
30-34	15	0	0
35-39	10	+1	10	+49
40-44	6	+2	12
45-49	9	+3	27
	N=60	+3	∑fd1 = 10

Mean = A.m+ \(\frac{\sum f d}{N}\) xi
= 32+\(\frac{10}{60}\) x 5
= 32+ \(\frac{10}{12}\)
= 3282

Median = L + \(\left(\frac{\frac{N}{2}-F}{F m}\right)\) x i

Where L = 29.5
N/2=30 ,F=20 ,Fm=15 ,¡=5
= 29.5+\(\left(\frac{30-20}{15}\right)\) x5
= 29.5+\(\frac{10}{15}\) x 5
= 29.5+\(\frac{50}{15}\)
= 29.5 + 3.3
= 32.8

Question 3.
Calculate the mean, median and mode of the following frequency distribution?

CL	F
20-21	1
18-19	2
16-17	3
14-15	5
12-13	3
10-11	3
8-9	2
6-7	1
	N= 20

Answer:

C.I.	Midpoint	F	X1	Fx1
20-21	20.5	1	3	3
18-19	18.5	2	2	4
16-17	16.5	3	1	3
14-15	14.5	5	0	0
12-13	12.5	3	-1	-3
10-II	10.5	3	-2	-6
8-9	8.5	2	-3	-6
8-7	6.5	1	-4	-4
	N=20		Fx’ = -9

Mean = A.M. + \(\frac{\sum F x^1}{N}\) x i

Where,   A.M. = Assumed Mean
∑ = Sum of
Fx¹ = Frequency x deviates of the midpoints of the class intervals from the assumed mean in terms of class interval unit.
¡ = size of the class interval
Here,  A.M. = 14.5
∑FX¹ = -9
i=2

Mean = A.M. + \(\frac{\sum F x^1}{N}\) x i
= 14.5+\(\frac{-9}{20}\) x 2
= 14.5 + -9
= 13.6

Median = L + \(\left(\frac{\frac{N}{2}-F}{F m}\right)\) x i
Where L = Lower limit of the class interval containing the Median i.e. \(\frac{n}{2}\)th score.
F = Cumulative frequency up to ‘L’
Fm = Frequency of the C.I. containing the Medium
i = size of the class interval

When counting from the top or bottom we found that \(\frac{n}{2}\)th score i.e. the 10th. score lies in the class interval 14-15.
∴ L = 13.5
F=9 Fm=5 ¡=2

Mdn = 13.5+\(\left(\frac{\frac{20}{2}-9}{5}\right)\) x2
=13.5+4=13.9
Mode = 3Mdn – 2 Mean
=3 x 13.9 – 2 X 13.6
=41.7 – 27.2
=14.5
∴ Calculated Mean = (3.6), Median = (13.9) and Mode = 14.5

Question 4.
Calculate the mean, and median of the following frequency distribution?

C.I.	F
90-94	1
85-89	3
80-84	5
75-79	7
70-74	9
65-69	6
60-64	4
55-59	3
50-54	2
	N = 40

Answer:

C.I.		Midpoint	F	X1	Fx1
90-94	92	1	4	4
85-89	87	3	3	9
80-84	82	5	2	10
75-79	77	7	1	7
70-74	72	9	0	0
65-69	67	6	-1	-6
60-64	62	4	-2	-8
55-59	57	3	-3	-9
50-54	52	2	-4	-8
		N=40		∑Fx’ = -l

Mean = A.M. + \(\frac{\sum F x^1}{N}\) x i
= 72+\(\frac{-1}{10}\) x 5
=72-125
=71.89

Median = L=69.5 L=69.5
N/20 =2O  N/20 =20
F= 15
Fm = 9
i = 5
= L+\(\left(\frac{\frac{\mathrm{N}}{2}-\mathrm{F}}{\mathrm{Fm}}\right)\) xi
= 65.5+\(\frac{(20-15) \times 5}{9}\)
= 65.9+\(\frac{25}{9}\)
= 69.5 + 2.77
= 72.2

Questions 5.
Find Mean in a short method?

Class interval	Frequencies
90-94	2
85-89	2
80-84	4
75-79	8
70-74	6
65-69	11
60-64	9
	N = 42

Answer:
Calculation means by the short method :

C.I.	(M)	F	(d1)	(Fd1)
90-94	92	2	+3	+6
85-89	87	2	+2	+4
80-84	82	4	+1	+4
75-79	77	8	0	0
70-74	77	6	-1	-6
65-69	67	11	-2	-22
60-64	62	9	-3	-27
		N=42		∑fd1= -41

Mean = A.M. + \(\frac{\sum F x^1}{N}\) x i
Where, A.M. = 77¹
∑fd¹ = -41
\(\frac{n}{2}\) = 21
i = 5

So, mean  77+\(\frac{-41}{42}\) x5
= 77+\(\left(\frac{-41}{42} \times 5\right)\)
= 77+\(\frac{205}{42}\)
= 77 + -4.8
= 72.7

questions 6.
Find out the Median of the scores?

C.I.	10-19	20-29	30-39	40-49	50-59	60-69
F	10	05	15	20	10	10

Answer:

Class interval	Frequency	Cum F.
10-19	10	10
20-29	05	15
30-39	15	30
40-49	20	50
50-59	10	60
60-69	10	70

Median = L+\(\left(\frac{\frac{N}{2}-F}{F m}\right)\) xi
L = Lower limit of the CI. in \(\frac{n}{2}\) F = The frequency below the CI. of \(\frac{n}{2}\)
Fm = The F at N/2 scores
i = class Interval
\(\frac{n}{2}\)= \(\frac{70}{2}\) = 35

L= 39.5   F = 30    F = 20  i =10
Median = L+\(\left(\frac{\frac{N}{2}-F}{F m}\right)\) xi
= 39.5+\(\left(\frac{35-30}{20}\right)\)x10
= 39.5+\(\left(\frac{35-30}{20}\right)\)x10
= 39.5+\(\frac{5}{20}\)x10
= 39.5+\(\frac{5}{2}\)x10
= 39.5 + 2.5
= 42.6

questions 7.

CL	F
30-34	2
25-29	3
20-24	3
15-19	10
10-14	4
5-9	5
0-4	2
	N =30

Answer:

CL	F	Cum F
30-34	2	30
25-29	3	28
20-24	3	25
15-19	10	19
10-14	4	9
5-9	5	5
0-4	2	2
	N =30

Here \(\frac{n}{2}\) =15
F=15(15-19) in the CI.
Fm=1O i =15
Median = L+\(\left(\frac{\frac{N}{2}-F}{F m}\right)\) xi
= 14.5+\(\frac{(15-19)}{10}\) x 5
= 14.5+\(\frac{4}{10}\) x5
= 14.5+2
= 16.5

Questions 8.
Calculate the Mean, Median and Mode from the class interval given?

CL	F
35-39	2
30-34	3
25-29	5
20-24	7
15-19	5
10-14	3
5-9	3
	N =28

Answer:

CL	F	D1	Fd1
35-39	2	+3	6	6
30-34	3	+2	5	+17
25-29	5	+1	5
20-24	7	0	0
15-19	5	-1	-5
10-14	3	-2	-6	-20
5-9	3	-3	-9
	N =28		∑fd1= -3

Mean=A.M. +C.I.
= 17+\(\frac{-3}{8}\) x5
= 17+\(\frac{-15}{28}\) x5
= 17-0.18
= 16.2

Median = L+\(\left(\frac{\frac{N}{2}-F}{F m}\right)\) xi
= 14.5+\(\frac{14-11}{7}\) x5
= 14.5+\(\frac{15}{7}\)
= 14.5+2.1
= 16.7

Mode = 3 Median -2 Mean
=3X 16.7- 2X 16.2
=50.1 -32.4
= 17.6
∴ Calculated Mean, Median and Mode are 16.2, 17.7 and 17.6 respectively

Questions 9.
Find out the Mean, Median and Mode by the long method?

C.I.	F
55-59	1
50-54	1
45-49	3
40-44	4
35-39	6
30-34	7
25-29	12
20-24	6
15-19	8
10-14	2
	N = 50

Answer:

C.I.	F	X	Fx
55-59	1	57	57
50-54	1	52	52
45-49	3	47	141
40-44	4	42	168
35-39	6	37	222
30-34	7	32	224
25-29	12	27	324
20-24	6	22	132
15-19	8	17
	N = 50		∑Fx1 = 1472

mean = \(\frac{\sum F x}{N}\)
= \(\frac{1472}{50}\)
= 29.4
Median = L+\(\left(\frac{\frac{N}{2}-F}{F m}\right)\) xi

Where L = 34.5
N/2 = 25
F = 35
Fm = 6
i = 5

So, Median = L+\(\left(\frac{\frac{N}{2}-F}{F m}\right)\) xi
= 34.5+\(\left(\frac{25-35}{6}\right)\) x5
= 34.5+\(\frac{-10}{6}\) x5
= 34.5+\(\frac{-50}{6}\)
= 34.5-8.3
= 26.2

Mode =3. Median -2 Mean
=3X 26.2 – 2X 29.4
= 76.6 – 58.8
= 19.5

Questions 10.
Find out the Mean, Median of the frequency distribution?

C.I.	F
20-21	1
18-19	2
16-17	3
14-15	5
12-13	3
10-11	3
8-9	2
6-7	1

Answer:

C.I.	Midpoint	F	X1	Fx1
20-21	20-5	1	3	3
18-19	18-5	2	2	4
16-17	16-5	3	1	3
14-15	14-5	5	0	0
12-13	12-5	3	-1	-3
10-11	10-5	3	-2	-6
8-9	8-5	2	-3	-6
6-7	6-5	1	-4	-4
		N = 20		∑fx1= -9

Mean = A.M. + \(\frac{\sum F x^1}{N}\) x i
where A.M, = Assumed mean
Σ=Sumtotal
Fx¹ = Frequency x deviations of the points of the class intervals from the assumed mean X¹ Terms of the class intervals.
i = size of the class intervals
Here, A.M. = 14.5
∑Fd =-9

Mean = 14.5+\(\frac{-9}{20}\) x2
= 14.5-9
= 13.6

Where L = Lower limit of the class interval containing the Median \(\frac{n}{2}\) score,
F = cumulative frequency up to L.

Fm = Frequency of the class interval containing the median.
i = size of the class interval
L= 13.5, f=9, Fm5, i=2
Median = 13.5+ \(\frac{10.9}{5}\) x2
= 13.5+\(\frac{1}{5}\)x2
= 13.5+4
= 13.9
Mode = 3 Median – 2 Mean
= 3 x 13.9 – 2 x 13.8
= 41.7 – 27.2
= 14.5

Question 11.
Find out the Mean and Median by the long method.

C.I.	F
10-14	4
15-19	5
20-24	9
25-29	18
30-34	11
35-39	5
40-44	6
45-49	3
50-54	1
	N = 60

Answer:

C.I.	X	F	FX
10-14	12	4	48
15-19	17	5	85
20-24	22	9	198
25-29	27	18	486
30-34	32	11	352
35-39	37	5	185
40-44	42	6	252
45-49	47	3	94
50-54	5	1	52
		N = 60	∑Fx = 1760

Mean = \(\frac{\sum F x}{N}\)
= \(\frac{1760}{60}\)
= 29.0

Calculation of Median;
Median = L+\(\left(\frac{\frac{N}{2}-F}{F m}\right)\) xi
Where
L = 24.5\(\frac{N}{2}\)
= \(\frac{60}{2}\)
= 30
Fm = 18,F = 17 ,i=5
So, Median = L+\(\left(\frac{\frac{N}{2}-F}{F m}\right)\) xi
= 24.5+\(\left(\frac{30-17}{18}\right)\) x5
= 24.5+3.6
= 28.0

Mode = 3 Median – 2 Mean
=3 X28-2X29=26.3

Question 12.
Find out the Mean of the following distribution?

C.I.	X
15-19	6
20-24	5
25-29	8
30-34	15
35-39	10
40-44	9
45-49	7
	N = 60

Answer:

C.I.	X	D1	FD1
15-19	6	-3	-18
20-24	5	-2	-10
25-29	8	-1	-8
30-34	15	0	0
35-39	10	1	10
40-44	9	2	18
45-49	7	3	21
	N = 60		∑Fd1= -13

Mean = A.M.+\(\frac{\sum F d^{\prime}}{N}\) x i
Suppose, the mean is within the 30-34 class interval
∴ A.M. = 32
In C.I. 25-29,
d = \(\frac{27-32}{5}\)
= \(\frac{-5}{6}\)
= -1

∴ Mean = A.M.+\(\frac{\sum F d^{\prime}}{N}\) x i
= 32+\(\frac{13}{60}\)x5
= 32+\(\frac{13}{12}\)
= 32+1.09
= 33.09

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