# Class Interval And Frequency Distribution Of Vendor Biology Essay

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In order to determine the frequency distribution of the above data, firstly we have to identify the class interval for this data. As the data given is in the decimal form, thus we need to follow the following steps to identify the class interval:

Step 1: To divide the whole data into class interval we need to identify the lower bound and upper bound. Lower bound and upper bound of the class interval can be calculated through following formula:

Step 2: In order to calculate the lower bound and the above bound, we need to identify the highest and the lowest value from the above data, the highest value is 103.7 and the lowest value is 96.8.

Step 3: For lower and upper bound we need to decide number of class interval or the number of classes we want in our distribution. Let us consider the number of classes as 5.

Step 4: On the basis of above data, we will calculate the upper bound and lower bound of our class interval:

In equation 1, we know that:

Highest value in the data= 103.7

Lowest value in the data= 96.8

Number of classes= 5

Thus Lower bound of class interval=

Thus Lower Bound= 1.38

Similarly from equation 2, we have:

Upper bound of class interval=

Thus Upper Bound= 1.725 ~ 1.72

Step 5: The upper and lower bound restricts the limit of class interval, the lower bound of the class interval is 1.38 and upper bound is 1.725, this suggest that the differences in the class interval should be more than 1.38 but should not be more than 1.725. Thus we will select 1.6 as the differences in class interval for our convenience.

Step 6: The first upper limit of the class interval is the largest number of that interval but is just less than the lower limit of second class interval. In the data given for the vendor 1, the number of class interval is 5 and thus the lower limit of each class interval can be obtained by adding 1.6 repeatedly to the lowest number of the whole data i.e. to 96. 8 up to the five classes say for e.g. 96.8 + 1.6= 98.4 is the lower limit of next class interval. Similarly, the lower limit of second class limit is just higher than the upper limit of first class i.e. 98.3 and thus upper limit of each interval can be obtained by adding 1.6 repeatedly to the 98.3. By applying this procedure, the class interval so obtained is given in Table 1:

Step 7: On the basis of the class intervals so identified, we can identify the data that lies in the particular class interval and put one tally marks for each number, the number of tallies form the number of frequency for that class interval; this can be shown thorough the following Table 1:

Table : Class interval and Frequency Distribution of Vendor 1:

## Class Interval

## Tallies

## Frequency

96.8-98.3

## â”‚â”‚â”‚â”‚ â”‚

6

98.4-99.9

## â”‚â”‚â”‚â”‚ â”‚â”‚â”‚â”‚ â”‚

11

100-101.5

## â”‚â”‚â”‚â”‚ â”‚

6

101.6-103.1

## â”‚

1

103.2-104.7

## â”‚

1

Total

26

## Calculation of Frequency Distribution for Vendor 2:

Similarly as above, we will calculate the frequency distribution for vendor 2:

## Vendor 2

106.8

106.8

104.7

104.7

108.0

102.2

103.2

103.7

106.8

105.1

104.0

106.2

102.6

100.3

104.0

107.0

104.3

105.8

104.0

106.3

102.2

102.8

104.2

103.4

104.6

103.5

106.3

109.2

107.2

105.4

106.4

106.8

104.1

107.7

107.7

In order to determine the frequency distribution of the above data, firstly we have to identify the class interval for this data. As the data given is in the decimal form, thus we need to follow the following steps to identify the class interval:

Step 1: To divide the whole data into class interval we need to identify the lower bound and upper bound. Lower bound and upper bound of the class interval can be calculated through following formula:

Step 2: In order to calculate the lower bound and the above bound, we need to identify the highest and the lowest value from the above data, the highest value is 109.2 and the lowest value is 100.3

Step 3: For lower and upper bound we need to decide number of class interval or the number of classes we want in our distribution. Let us consider the number of classes as 5.

Step 4: On the basis of above data, we will calculate the upper bound and lower bound of our class interval:

In equation 1, we know that:

Highest value in the data= 109.2

Lowest value in the data= 100.3

Number of classes= 5

Thus Lower bound of class interval=

Thus Lower Bound= 1.78

Similarly from equation 2, we have:

Upper bound of class interval=

Thus Upper Bound= 2.225

Step 5: The upper and lower bound restricts the limit of class interval, the lower bound of the class interval is 1.78 and upper bound is 2.225, this suggest that the differences in the class interval should be more than 1.78 but should not be more than 2.225. Thus we will select 2 as the differences in class interval for our convenience.

Step 6: The first upper limit of the class interval is the largest number of that interval but is just less than the lower limit of second class interval. In the data given for the vendor 1, the number of class interval is 5 and thus the lower limit of each class interval can be obtained by adding 2 repeatedly to the lowest number of the whole data i.e. to 100.3 up to the five classes say for e.g. 100.3 + 2 = 102.3 is the lower limit of next class interval. Similarly, the lower limit of second class limit is just higher than the upper limit of first class i.e. 102.2 and thus upper limit of each interval can be obtained by adding 2 repeatedly to the 102.2. By applying this procedure, the class interval so obtained is given in Table 2:

Step 7: On the basis of the class intervals so identified, we can identify the data that lies in the particular class interval and put one tally marks for each number, the number of tallies form the number of frequency for that class interval; this can be shown thorough the following Table 2:

Table : Class interval and Frequency Distribution of Vendor 2:

## Class Interval

## Tallies

## Frequency

100.3-102.2

## â”‚â”‚â”‚

3

102.3-104.2

## â”‚â”‚â”‚â”‚ â”‚â”‚â”‚â”‚ â”‚

11

104.3-106.2

## â”‚â”‚â”‚â”‚ â”‚â”‚â”‚

8

106.3-108.2

## â”‚â”‚â”‚â”‚ â”‚â”‚â”‚â”‚ â”‚â”‚

12

108.3-110.2

## â”‚

1

Total

35

## Solution 2:

## Calculation for the Mean of resistors provided by Vendor 1:

## Vendor 1

96.8

100.0

100.3

98.5

98.3

98.2

99.6

99.4

99.9

101.1

103.7

97.7

99.7

101.1

97.7

98.6

101.9

101.0

99.4

99.8

99.1

99.6

101.2

98.2

98.6

Calculation of Mean (la moyenne ) for Vendor 1 by Simple Method

Mean = Total value of Resistance of Resistors / Total number of Observed Resistance

= 96.8 + 100.0 + 100.3 + 98.5 + 98.3 + 98.2 + 99.6 + 99.4 + 99.9 + 101.1 + 103.7 + 97.7 +99.7 + 101.1 + 97.7 + 98.6 + 101.9 + 101.0 + 99.4 + 99.8 +99.1 + 99.6 + 101.2 + 98.2 + 98.6 / 25.

= 2489.4 / 25 = 99.576

Mean = 99.576

Or

## Calculation of Mean using Class Intervals:

## S.No

## Class Intervals

## Mid Point (x)

## Frequency (f)

## (xf)

1.

96.8-98.3

97.55

6

585.30

2.

98.4-99.9

99.15

11

1090.65

3.

100-101.5

100.75

6

604.50

4.

101.6-103.1

102.35

1

102.35

5.

103.2-104.7

103.95

1

103.95

Total

25

2486.75

Mean (la moyenne ) = 2486.75 / 25

Mean (la moyenne ) = 99.47

## Working Notes:

## Calculation of Mid Points:

## Formula:

Class Interval = 96.8-98.3

So, Mid Point = 96.8 + 98.3 / 2 = 97.55

Class Interval = 98.4-99.9

So, Mid Point = 98.4 + 99.9 / 2 = 99.15

Class Interval = 100-101.5

So, Mid Point = 100.0 + 101.50 / 2 = 100.75

Class Interval = 101.6-103.1

So, Mid Point = 101.6 + 103.1 / 2 = 102.35

Class Interval = 103.2-104.7

So, Mid Point = 103.2 + 104.7 / 2 = 103.95

## Calculation of Product of Mid Point (x) and Frequency (f) i.e. (xf):

Mid Point (x) = 97.55, Frequency (f) = 6

(xf) = 97.55 * 6 = 585.30

Mid Point (x) = 99.15, Frequency (f) = 11

(xf) = 99.15 * 11 = 1090.65

Mid Point (x) = 100.75, Frequency (f) = 6

(xf) = 100.75 * 6 = 604.50

Mid Point (x) = 102.35, Frequency (f) = 1

(xf) = 102.35 * 1 = 102.35

Mid Point (x) = 103.95, Frequency (f) = 1

(xf) = 103.95 * 1 = 103.95

## Calculation of Summation of (xf) i.e. ∑ xf:

+ 1090.65 + 604.50 + 102.35 + 103.95 = 2486.75

## Calculation of Summation of (f) i.e. ∑ f:

6 + 11 + 6 + 1 + 1 = 25

## Calculation for the mean of resistors provided by Vendor 2:

## Vendor 2

106.8

106.8

104.7

104.7

108.0

102.2

103.2

103.7

106.8

105.1

104.0

106.2

102.6

100.3

104.0

107.0

104.3

105.8

104.0

106.3

102.2

102.8

104.2

103.4

104.6

103.5

106.3

109.2

107.2

105.4

106.4

106.8

104.1

107.7

107.7

Calculation of Mean (la moyenne ) for Vendor 2 by Simple Method:

## Mean = Total value of Resistance of Resistors / Total number of Observed Resistance

= 106.8 + 106.8 + 104.7 + 104.7 + 108.0 + 102.2 + 103.2 + 103.7 + 106.8 + 105.1 + 104.0 + 106.2 + 102.6 + 100.3 + 104.0 + 107.0 + 104.3 + 105.8 + 104.0 + 106.3 + 102.2 + 102.8 + 104.2 + 103.4 + 104.6 + 103.5 + 106.3 + 109.2 + 107.2 + 105.4 + 106.4 + 106.8 + 104.1 + 107.7 + 107.7 = 3678 / 35.

= 105.08

## Mean = 105.08

Or

## Calculation of Mean using Class Intervals:

## Class Intervals

## Mid Point (x)

## Frequency (f)

## (xf)

100.3-102.2

101.25

3

303.25

102.3-104.2

103.25

11

1135.75

104.3-106.2

105.25

8

842

106.3-108.2

107.25

12

1287

108.3-110.2

109.25

1

109.25

## Total

## 35

## 3677.25

Mean (la moyenne ) = 3677.25 / 35

Mean (la moyenne ) = 105.06

## Working Notes:

## Calculation of Mid Points:

## Formula:

Class Interval = 100.3-102.2

So, Mid Point = 100.3 + 102.2 / 2 = 101.25

Class Interval = 102.3-104.2

So, Mid Point = 102.3 + 104.2 / 2 = 103.25

Class Interval = 104.3-106.2

So, Mid Point = 104.3 + 106.2 / 2 = 105.25

Class Interval = 106.3-108.2

So, Mid Point = 106.3 + 108.2 / 2 = 107.25

Class Interval = 108.3-110.2

So, Mid Point = 108.3 + 110.2 / 2 = 109.25

## Calculation of Product of Mid Point (x) and Frequency (f) i.e. (xf):

Mid Point (x) = 101.25, Frequency (f) = 3

(xf) = 101.25* 3 = 303.25

Mid Point (x) = 103.25, Frequency (f) = 11

(xf) = 103.25* 11 = 1135.75

Mid Point (x) = 105.25, Frequency (f) = 8

(xf) = 105.25* 8 = 842

Mid Point (x) = 107.25, Frequency (f) = 12

(xf) = 107.25* 12 = 1287

Mid Point (x) = 109.25, Frequency (f) = 1

(xf) = 109.25 * 1 = 109.25

## Calculation of Summation of (xf) i.e. ∑ xf:

303.25 + 1135.75 + 842 + 1287 + 109.25 = 3677.25

## Calculation of Summation of (f) i.e. ∑ f:

3 + 11 + 8 + 12 + 1 = 35

## Solution 3:

In order to calculate the similarity and differences in the variability of both the vendor, first we need to calculate the variance for both the vendors. Variability or variance is the square of standard deviation for any data and thus to calculate variance we need to calculate the standard deviation for both the resistors. The variance and standard deviation for the class intervals is given by the formula as follows:

Where X= Midpoints of the class interval

= Mean for the resistors of Vendor 1

f= Frequency

N= Total number of resistors

Thus for calculating the standard deviation and variance for both the resistors, we will required to use the mean for both the vendors. The mean for both the vendors have been calculated in the above solution 2 and thus we will use the table 3 and 4 for the calculation of variability:

## Calculation of variance for Vendor 1:

Table : Calculation or standard deviation of Vendor 1:

## Class Interval

## Mid Point (x)

## Frequency (f)

96.8-98.3

97.55

6

-1.92

3.69

22.14

98.4-99.9

99.15

11

-0.32

.1024

1.1264

100-101.5

100.75

6

1.28

1.64

9.84

101.6-103.1

102.35

1

2.88

8.29

8.29

103.2-104.7

103.95

1

4.48

20.07

20.07

Total

25

61.4664

Mean for the vendor 1 = 99.47 (from solution 2)

From the above table we have:

From equation 3 we have:

Thus standard deviation for vendor 1= 1.568

Variance as given by equation 4 is square of standard deviation and thus

Variance for vendor 1=

= 2.458~2.46

## Calculation of variance for Vendor 2:

Table 4: Calculation or standard deviation of Vendor 2:

## Class Interval

## Mid Point (x)

## Frequency (f)

100.3-102.2

101.25

3

-3.81

14.52

43.56

102.3-104.2

103.25

11

-1.81

3.28

36.08

104.3-106.2

105.25

8

0.19

0.036

.288

106.3-108.2

107.25

12

2.19

4.79

57.48

108.3-110.2

109.25

1

4.19

17.55

17.55

Total

35

154.958

Mean for the vendor 2 = 105.06 (from solution 2)

From the above table 4 we have:

From equation 3 we have:

Thus standard deviation for vendor 2 = 2.104

Variance as given by equation 4 is square of standard deviation and thus

Variance for vendor 1=

= 4.427

From the above calculation it is clear that the variance of vendor 1 is 2.46 where as the variance of vendor 2 from the mean is 4.427 and thus vendor 2 is more variable as compared to the vendor 1. It also shows that there the variability of both the vendors are different from each other (Variance and standard deviation).

## Solution 4:

## Graph for the vendor 1:

The above graph shows the resistance of the resistors provided by the vendor 1 to the procurement specialist. The above graph depicts that the resistance of most of the resistors lies between the ranges of 98 to 102. Only the resistance of approximately four resistors lies either above or below this limit. This shows that the resistance of most of the resistors provided by the vendor 1 is consistently same with slight differences which indicate that strong efforts are made to maintain the quality. The above graph shows that the data is consistent and lies between mean ranges of 100. Thus the resistors of vendor 1 have consistently same resistance with slight difference.

## Solution 5:

## Graph for the vendor 2:

The above graph shows that the resistance of the resistors provided by vendor 2 to the purchase specialist. It depicts that resistance of the resistors fluctuate a lot with each other as it shows much differences. Also the graph shows that the resistors so provided by the vendor 2 have very high resistance as compared to the resistors of vendor 2. The resistance of all the resistors is above 100 which indicate that the resistors have the high resistance. The resistances of the resistors are not consistently same as compared to the resistors of vendor 1 which indicate that vendor 2 is lacking in terms of quality control.

## Solution 6:

The confidence interval provides an estimated range of the values which is possibly to include the unknown population parameter. It is the estimated range which is calculated for the given sample of data. In the case of vendor 1 and vendor 2 we have mean values and standard deviation and thus we will calculate the confidence interval for the mean of both the vendors. In order to calculate the confidence interval of both the vendors, we need to follow some steps which are explained as under:

## Calculation for the Confidence Interval of Vendor 1:

To calculate the confidence interval following steps should be followed:

The first step measures the sample statistics which will help to analyze the data and helps in the calculation of confidence interval. In the case of vendor1 the sample statistics is as follows:

Mean = 99.47

Standard deviation = 1.568

Variance = 2.46

The second step selects the confidence level for the range of confidence interval. As the value of α is given as 5%, the level of confidence can be calculated with the help of following formula:

Therefore;

Thus confidence level = 95 % (What is a Confidence Interval?)

On the basis of above confidence level, the confidence interval can be calculated through following formula:

By the empirical rule; 95 % of the time falls in the interval of

Here µ=

Thus confidence interval for the mean of the resistors provided by vendor 1 can be calculated as follows:

The lower range of confidence interval for the vendor 1 can be calculated as:

= 98.85

Thus the lower limit of confidence interval for the mean of the resistors provided by vendor 1 is 98.85.

The upper limit of the confidence interval is as follows:

= 100.08

Thus we are 95 % confident that the mean of the resistors provided by vendor 1 lies in the range of 98.85 to 100.08, and the confidence interval for the vendor 1 is ± 0.61 (Confidence Intervals for Means).

## Calculation for the Confidence Interval of Vendor 2:

To calculate the confidence interval following steps should be followed:

The first step measures the sample statistics which will help to analyze the data and helps in the calculation of confidence interval. In the case of vendor 2 the sample statistics is as follows:

Mean = 105.06

Standard deviation = 2.104

Variance = 4.427

The second step selects the confidence level for the range of confidence interval. As the value of α is given as 5%, the level of confidence can be calculated with the help of following formula:

Therefore;

Thus confidence level = 95 % (What is a Confidence Interval?)

On the basis of above confidence level, the confidence interval can be calculated through following formula:

By the empirical rule; 95 % of the time falls in the interval of

Here µ=

Thus confidence interval for the mean of the resistors provided by vendor 2 can be calculated as follows:

The lower range of confidence interval for the vendor 2 can be calculated as:

= 104.38

Thus the lower limit of confidence interval for the mean of the resistors provided by vendor 2 is 104.38 (Confidence Intervals for Means).

The upper limit of the confidence interval is as follows:

= 105.78

Thus we are 95 % confident that the mean of the resistors provided by vendor 2 lies in the range of 104.38 to 105.78, and the confidence interval for the vendor 2 is ± 0.7.

The confidence interval of vendor 1 is ±.61 and that of vendor 2 is ±0.7 which indicates that the resistance of most of the resistors provided by vendor 1 is closer to the mean value as compared to that of the resistors provided by vendor 2. The confidence interval of vendor 1 shows that the resistance of the resistors is more normal as compared to the resistance of resistors provided by vendor 2. The mean of resistance of resistors of vendor 2 lies in the upper range of confidence interval which shows that the resistance of resistors is not normal and consistent and it have high resistance. In contrast to vendor 2, the mean of the resistors provided by Vendor 1 lies in between the range of confidence interval which shows that the resistors have normal resistance and provide consistent performance.

## Solution 7:

All the statistical information so calculated and derived from all the measurements suggests that the Vendor 1 provided more qualitative resistors as compared to Vendor 2 and thus Vendor 1 is recommended for the purchasing of resistors to the procurement specialist. Vendor 1 is recommended for the purchase of the resistors because of the following reasons:

The mean of resistors provided by vendor 1 lies between the range of medium resistance and shows the nature of all the resistors as the resistance of all the resistors lies between the ranges of 98 to 102 which shows that much concentration is given to the maintenance of high quality so that all the resistors have a consistent performance. Against this, the resistors provided by the Vendor 2 have very high resistance as shown by its mean; also the mean does not show the nature of all the resistors. The resistance of all the resistors so provided by vendor 2 fluctuates with each other which show that they do not provide consistently same performance.

The standard deviation and variance of vendor 1 is less than that of vendor 2 which shows that the resistance of the resistors provided by vendor 2 is more variable as compared to vendor 1 and thus vendor 1 is recommended over vendor 2 as they provide consistent and high quality resistors with less variability and standard deviation from the mean.

The graph of the vendor 1 shows that the resistance of the entire resistor is consistent same and lies in the controlling range of 96 to 102 with less variability. Against this, the graph of vendor 2 shows that the resistances of resistors fluctuate with each other and does not shows consistency and thus vendor 1 is recommended.

The confidence interval of vendor 1 shows that resistance of the resistors is more normal and consistent as compared to vendor 2 and thus vendor 1 is recommended over vendor 2.

All the analysis done above shows that vendor 1 provides high quality and consistent product as compared to vendor 2 and thus vendor 1 is recommended.

## References:

Confidence Intervals for Means. (n.d.). Retrieved August 21, 2012, from http://www.stat.wmich.edu: http://www.stat.wmich.edu/s160/book/node46.html

Quantitative Frequency Distributions and Histograms. (n.d.). Retrieved August 21, 2012, from http://www2.selu.edu: http://www2.selu.edu/Academics/Faculty/dgurney/Math241/StatTopics/HistGen.htm

Variance and standard deviation. (n.d.). Retrieved August 21, 2012, from http://www.statcan.gc.ca: http://www.statcan.gc.ca/edu/power-pouvoir/ch12/5214891-eng.htm

What is a Confidence Interval? (n.d.). Retrieved August 21, 2012, from http://stattrek.com: http://stattrek.com/estimation/confidence-interval.aspx