Assignment of UNIT -1

(SM)      Date : 27/8/2010

Assignment of Unit -1 ( Submission Date: 13/9/2010)

Q1. The following data represent the weights (in kilograms) of a sample of 25 horses:

164 148 137 157 173 156 177 172 169 165
145 168 163 162 174 152 156 168 154 151
174 146 134 140 171

Use these data to answer the following questions:
a. Construct a stem-and-leaf display for the weights.

Q2. A study was carried out on the degree of job satisfaction among doctors and nurses in rural and urban areas. To describe the sample a cross-tabulation was constructed which included the sex and the residence (rural or urban) of the doctors and nurses interviewed. This was useful because in the analysis the opinions of male and female staff had to be compared separately for rural and urban areas. Construct Cross Tabulation

Q3.   These are the marks of two judges for ten entries in a cookery competition

Entry A B C D E F G H I J

Judge 1  63 84 31 47 59 72 80 93 38 86

Judge 2  51 77 36 45 60 64 72 75 43 85

Draw a scatter diagram and the line of best fit

Q4. Find the Mean , median, Mode from the  following numbers:36, 38, 33, 34, 32, 30, 34, 35.

Q5. What are the quartiles and 60th percentile for the data given here? 5,6,6,7,7,8,8,8,9,10.?

Q6. Find out the range of the data 5700 and 9000 ?

Q7.Find Q1, Q2, Q3 from the data 23 25 29  31 34 35  36 37 38  40 42.

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Q8. Find the sample variance and population variance and standard deviation , coefficient of variation   of the data  32 50 40  30 ?

Q9. Find out the z-score for the following data 93,119,101,77,110,95,90,91,102,122.

Q10. Find out the five number summary 50,57,41,54,54,38,63,48,59,46.

Q11. Find out weighted mean from 2,6,4,4,2,2 and 1200,1500,4000,2500,2340,2345.

Q12.  Find out sample variance , population variance from the data

Data Entry Frequency 100 8 150 15 200 21 250 14 300 5

Bold is frequency

Q13. Draw the histogram , ogive from the above data.

Q14. Difference between Qualitative and Quantitative.

Q15. Write all  the Formula of unit -1

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Assignment of UNIT -1

(SM)                                                                      Date : 27/8/2010

Assignment of Unit -1                             Submission Date: 13/9/2010

Q1. The following data represent the weights (in kilograms) of a sample of 25 horses:

164 148 137 157 173 156 177 172 169 165
145 168 163 162 174 152 156 168 154 151
174 146 134 140 171

Use these data to answer the following questions:
a. Construct a stem-and-leaf display for the weights.

Q2. A study was carried out on the degree of job satisfaction among doctors and nurses in rural and urban areas. To describe the sample a cross-tabulation was constructed which included the sex and the residence (rural or urban) of the doctors and nurses interviewed. This was useful because in the analysis the opinions of male and female staff had to be compared separately for rural and urban areas. Construct Cross Tabulation

Q3.   These are the marks of two judges for ten entries in a cookery competition

Entry A B C D E F G H I J

Judge 1  63 84 31 47 59 72 80 93 38 86

Judge 2  51 77 36 45 60 64 72 75 43 85

Draw a scatter diagram and the line of best fit

Q4. Find the Mean , median, Mode from the  following numbers:36, 38, 33, 34, 32, 30, 34, 35.

Q5. What are the quartiles and 60th percentile for the data given here? 5,6,6,7,7,8,8,8,9,10.?

Q6. Find out the range of the data 5700 and 9000 ?

Q7.Find Q1, Q2, Q3 from the data 23 25 29  31 34 35  36 37 38  40 42.

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Q8. Find the sample variance and population variance and standard deviation , coefficient of variation   of the data  32 50 40  30 ?

Q9. Find out the z-score for the following data 93,119,101,77,110,95,90,91,102,122.

Q10. Find out the five number summary 50,57,41,54,54,38,63,48,59,46.

Q11. Find out weighted mean from 2,6,4,4,2,2 and 1200,1500,4000,2500,2340,2345.

Q12.  Find out sample variance , population variance from the data

Data Entry Frequency 100 8 150 15 200 21 250 14 300 5

Bold is frequency

Q13. Draw the histogram , ogive from the above data.

Q14. Difference between Qualitative and Quantitative.

Q15. Write all  the Formula of unit -1

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Assignment of UNIT -1

(SM)                                                                      Date : 27/8/2010

Assignment of Unit -1

Q1. The following data represent the weights (in kilograms) of a sample of 25 horses:

164 148 137 157 173 156 177 172 169 165
145 168 163 162 174 152 156 168 154 151
174 146 134 140 171

Use these data to answer the following questions:
a. Construct a stem-and-leaf display for the weights.

Q2. A study was carried out on the degree of job satisfaction among doctors and nurses in rural and urban areas. To describe the sample a cross-tabulation was constructed which included the sex and the residence (rural or urban) of the doctors and nurses interviewed. This was useful because in the analysis the opinions of male and female staff had to be compared separately for rural and urban areas. Construct Cross Tabulation

Q3.   These are the marks of two judges for ten entries in a cookery competition

Entry A B C D E F G H I J

Judge 1  63 84 31 47 59 72 80 93 38 86

Judge 2  51 77 36 45 60 64 72 75 43 85

Draw a scatter diagram and the line of best fit

Q4. Find the Mean , median, Mode from the  following numbers:36, 38, 33, 34, 32, 30, 34, 35.

Q5. What are the quartiles and 60th percentile for the data given here? 5,6,6,7,7,8,8,8,9,10.?

Q6. Find out the range of the data 5700 and 9000 ?

Q7.Find Q1, Q2, Q3 from the data 23 25 29  31 34 35  36 37 38  40 42.

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Q8. Find the sample variance and population variance and standard deviation , coefficient of variation   of the data  32 50 40  30 ?

Q9. Find out the z-score for the following data 93,119,101,77,110,95,90,91,102,122.

Q10. Find out the five number summary 50,57,41,54,54,38,63,48,59,46.

Q11. Find out weighted mean from 2,6,4,4,2,2 and 1200,1500,4000,2500,2340,2345.

Q12.  Find out sample variance , population variance from the data

Data Entry Frequency 100 8 150 15 200 21 250 14 300 5

Bold is frequency

Q13. Draw the histogram , ogive from the above data.

Q14. Difference between Qualitative and Quantitative.

Q15. Write all  the Formula of unit -1

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Summarization of Qualitative data & Quantitative

Qualitative data

data is a categorical measurement expressed not in terms of numbers, but rather by means of a natural language description. In statistics, it is often used interchangeably with "categorical" data. 

For example: favourite colour = "blue"  height = "tall"

Although we may have categories, the categories may have a structure to them. When there is not a natural ordering of the categories, we call these nominal categories. Examples might be gender, race, religion, or sport.
When the categories may be ordered, these are called ordinal variables. Categorical variables that judge size (small, medium, large, etc.) are ordinal variables. Attitudes (strongly disagree, disagree, neutral, agree, strongly agree) are also ordinal variables, however we may not know which value is the best or worst of these issues. Note that the distance between these categories is not something we can measure.

Quantitative data

data is a numerical measurement expressed not by means of a natural language description, but rather in terms of numbers. However, not all numbers are continuous and measurable. For example, the social security number is a number, but not something that one can add or subtract. 

 

For example: favourite colour = "450 nm" height = "1.8 m"

Quantitative data always are associated with a scale measure.Probably the most common scale type is the ratio-scale.

A more general quantitative measure is the interval scale. Interval scales also have a equidistant measure.

 

 

Standard Deviation

Standard Deviation and Variance

A commonly used measure of dispersion is the standard deviation, which is simply the square root of the variance. The variance of a data set is calculated by taking the arithmetic mean of the squared differences between each value and the mean value. Squaring the difference has at least three advantages:

1. Squaring makes each term positive so that values above the mean do not cancel values below the mean.
2. Squaring adds more weighting to the larger differences, and in many cases this extra weighting is appropriate since points further from the mean may be more significant.
3. The mathematics are relatively manageable when using this measure in subsequent statisitical calculations.

Because the differences are squared, the units of variance are not the same as the units of the data. Therefore, the standard deviation is reported as the square root of the variance and the units then correspond to those of the data set.
The calculation and notation of the variance and standard deviation depends on whether we are considering the entire population or a sample set. Following the general convention of using Greek characters to express population parameters and Arabic characters to express sample statistics, the notation for standard deviation and variance is as follows:
  =  population standard deviation
  =  population variance
s  =  estimate of population standard deviation based on sampled data
s²  =  estimate of population variance based on sampled data
The population variance is defined as:

=

The population standard deviation is the square root of this value.
The variance of a sampled subset of observations is calculated in a similar manner, using the appropriate notation for sample mean and number of observations. However, while the sample mean is an unbiased estimator of the population mean, the same is not true for the sample variance if it is calculated in the same manner as the population variance. If one took all possible samples of n members and calculated the sample variance of each combination using n in the denominator and averaged the results, the value would not be equal to the true value of the population variance; that is, it would be biased. This bias can be corrected by using ( n - 1 ) in the denominator instead of just n, in which case the sample variance becomes an unbiased estimator of the population variance.
This corrected sample variance is defined as:

s2    =

The sample standard deviation is the square root of this value.
Standard deviation and variance are commonly used measures of dispersion. Additional measures include the range and average deviation.

Assignment of UNIT -1(Part -AM) Short-cut Method

Assignment Submission Date : 11/8/2010


Using Short-cut Method

1.) Calculate step deviation for the following data:
Farm Size (in acres)   No. of farms
0— 40                    394
40— 80                  461
80—120                 391
120—160               334
160—200               169
200—240               113
240 and over


2.) Calculate step deviation

Variable      Frequency     Variable       Frequency
20—29         306              50—59          96
30—39         182              60—69          42
40—49          144             70—79          24



3.) From the following distribution, calculate the value of them MEAN

X               F        X            F
300—399 14 800— 899   62
400—499 46 900—999    48
500—599 58 1000--1099 22
600—699 76 1100—1199 6
700—799 68


4.)
Following are the monthly sales of a firm in a year:

Months                       1    2   3    4    5    6    7   8   9   10     11      12

Sales (in'000):         50 30  25  44  48  51  55  60 42  35  28     40

Find out mean deviation from the above data.

5.) Calculate the Mean Deviation from the following data relating to heights (to the nearest inch) of 100 children:

Height (inches): 60 61 62 63 64 65 66 67 68

No. of children: 2 0 15 29 25 12 10 4 3


6.) The following table gives the distribution of monthly wages of 1000 workers of a factory:
Wages ($)
No. of workers’ Wages ($) No. of workers
20 3 140 204
40 13 160 139
60 43 180 69
80 102 200 25
100 175 220 6
120 220 240 1

Find the mean deviation of the above group and also compute die dispersion.



7. Calculate the value of coefficient of mean deviation  of the following data:

Marks: 10—20 20—30 30—40 40—50 50—60 60—70 70—80 80—90

No. of Students: 2 6 12 18 25 20 10 7



8. Calculate  Mean Deviation for the following frequency distribution.

Age (years)	No. of persons	Age (years)	No. of persons
1-5	7	26—30	18
6—10	10	31—35	10
11—15	16	36—40	5
16—20	32	41—45	1
21—25	24

                

9.Find out coefficient of Mean deviation by using mean (X) from

The following data:

Class:	0—3	3—6	6-9	9-12	12—15	15—18	18—21
Frequency:	2	7	10	12	9	6	4

10.Find the mean deviation

Size	Frequency
1 and up to 10	1
1 and up to 20	3
1 and up to 30	6
1 and up to 40	8
1 and up to 50	10

Example of Short - cut Method (in AM)

Using The Short-Cut Method

Example:

Calculate the Mean Deviation from the data given in example, by using the short-cut method and assuming the mean value as 25 marks:

Solution:             

     

                   

Marks (X)	Mid Values (w.v) (ii)	No.of  students (f) (iii)	Dev. from assumed mean 25 (dx) (ignoring + and –sings) (iv)	Step Dev dx/I (v)	Total dev.from assumed Av. fdx. (vi)
0-10	5	6	20	2	12
10-20	15	5	10	1	5
20—30	25	8	0	0	0
30-40	35	15	10	1	15
40-50	45	7	20	2	14
50-60	55	6	30	3	18
60-70	65	3	40	4	12
		∑f = 50			76X10 = 760

Total Deviations from assumed mean (+ & — signs ignored) = 760

Note:  Where actual and assumed averages are in different class intervals a special adjustment is necessary. In such cases the frequency of the class in which the actual mean lies is treated separately. It is multiplied by the difference of the deviations of the mid-value from the actual and assumed averages. The product so obtained is subtracted from the total deviation from the assumed mean.

(i)    Now the number of items smaller than the actual mean (33.4 as calculated in example 15) = 19 (frequency of the mean class being ignored).

(ii)    Number of items bigger than the actual mean = 16 (frequency of the mean class being ignored).

(iii)    Frequency of the Mean class = 15

(iv)    Difference between Actual and Assumed means = (33.4-25) =   8.4

(v)    Difference of Deviations of mid-value of mean class from the actual and assumed average

= (35-33.4) — (35—25)

= 1.6—10 = 8.4 (+ and—signs ignored)

(vi)    Total Deviation from actual Mean

= 760+ (19 x 8.4) — (16 X 8.4) — (15 x 8.4) = 659.2

(vii)     Mean Deviation = 569.2/50 = 13.18 marks

Thus the answer derived in this manner is the same as we got in example No. 14.

It should be noted that the difference between the values of actual and assumed mean (8.4 in this case) and the 'difference of deviations, of mid-values of mean class from the actual and assumed average' [(32—33.4) (35—25) = 8.41 need not necessarily be the same figure. It is a coincidence that the two values are the same in the problem.


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Short-cut Method

While calculating the mean deviation from the mean it may be more convenient to use a short-cut method by assuming an arithmetic average. The process of calculating mean deviation by the short-cut method involves the following steps:—

(1)       Deviations of items are taken from an assumed mean and multiplied by their respective frequencies and the products so obtained are totaled.

(2)       Number of items less than the actual arithmetic average are multiplied by the difference between the actual and the assumed mean.

(3)       Similarly, the number of items more than the arithmetic average are multiplied by the difference between actual and the assumed mean.

(4)       The latter (No. 3) is deducted from the former (No. 2) and the balance is added to the sum of the products of deviations from the assumed mean and their frequencies (No. 1).

(5)       The resulting figure is divided by the number of items and it is the value of the mean deviation.

Example  is solved below by this short-cut method.

Marks (X)	Step Deviation From as. Av (25) (f)	No. of  Student’s (f)  (fd)	Number of  (f)	Dev. From as. Av. 25 (fd) (+ & - signs    (with + & -signs)
5	—2	5	10	8	—10
15	—1	8	15	0	—8
15	0	15	0	—8	30
35	+1	16	16		+16
45	+2	6	+12	12	+ 12
4	17	∑f= 50	∑fd’ = 46		∑fd = +10

Actual arithmetic average = 25 + (10/50×10) = 27


Total deviations from assumed average of 25 = 46 x 10=460

Adjustments


Number of items less than the actual arithmetic average (27) =28



Number of items more than the actual arithmetic average (27) =22


Difference between actual and assumed average =2


Total deviations from actual average or (∑fdX)


= 460 + (28 x2) — (22 x2)


= 460 + 56 — 44=472


    Mean Deviation         = (∑|fdX)/N = 472/50


                    =    9.44 marks.


(3)           Calculation of mean deviation in continuous series. The calculation of mean deviation in continuous series is done by the same procedure by which it is done in discrete series. In the short-cut method also the same procedure is -followed provided the assumed mean or median is in the same class-interval in which the actual mean or median is. If the assumed average is in a different class interval, further adjustments are necessary. 

Assignment UNIT -1

Submission Date: 6/8/2010

Assignment of MEAN

1. Twelve students were given a math test, and the times (in minutes) to complete it are listed below. Find the range of these times.


10, 9, 12, 11, 8, 15, 9, 7, 8, 6, 12, 10



2. A relay race was completed by 7 participants, and their race times are given below (in seconds). What is the range of race times?

13.2, 14.5, 12.9, 13.9, 15.6, 14.1, 12.3



3. What is the mean test score for the 6 scores listed below?

89, 93, 87, 86, 85, 94



4. The average annual wind speed for the 5 windiest cities in the U.S. is given below in miles per hour. What is the mean of these annual wind speeds?

15.4, 14.0, 13.5, 13.1, 12.9



5.Calculate arithmetic mean from the following data :
Marks (less than) : 80 70 60 50 40 30 20 10

No. of students : 100 90 80 60 32 20 13 5



6.The monthly profits in $ of 100, shops are distributed as follows :

Profit per shop: 0—100 0—200 0—300 0—400 0—500 0—600

No. of shop : 12 30 57 77 94 100



7.Income groups : 150—300 300—500 500—800 800—1200 1200—1800
   No. of firms :     40 32 26 28 42
   Average No .: of workers 8 12 7.5 8.5 4

Find the average salary paid in the whole market.

8.Find the missing frequency from the following data :


Marks  Frequency

5—10   12

10—15 16

15—20  5

20—25  14

25—30  10

30—35  8





9. The following is the frequency distribution of the marks obtained by 250 students in an examination. Compute the mean, the median and the mode.


Marks obtained No. of students Marks obtained No. of students

0—10     15  40—50  12

10—20   20  50—60  31

20—30   24  60—70  71

30-40     24  70—80  52