• September 9, 2016
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This is a solution of Unit 2 Business Decision Making Assignment Help that describes about Developing business

Data given: –

 Amount Spent (£) No. of customers 10-20 18 20-30 20 30-40 16 40-50 14 50-60 12 60-70 8 70-80 6 80-90 4 90-100 2

### i. Mean, Median and Mode (AC 2.1 & AC 2.2)

Mean:

 Amount Spent (£) No. of customers Mid-value fx 10 – 20 18 15 270 20 – 30 20 25 500 30 – 40 16 35 560 40 – 50 14 45 630 50 – 60 12 55 660 60 – 70 8 65 520 70 – 80 6 75 450 80 – 90 4 85 340 90 – 100 2 95 190 100 4120

(Marković, et. al., 2013)

Analysis:

Mean is termed as the average of the total data and get utilised as a base for decision making. In order to calculate the mean of above table the total sum of given data get divided by total number of customers such as:

Mean = 4120/ 100 = 41.20

£41.20 is the mean for the above table.(Marković, et. al., 2013)

Median:

 Amount Spent (£) No. of customers Cumulative 20-Oct 18 18 20-30 20 38 30-40 16 54 40-50 14 68 50-60 12 80 60-70 8 88 70-80 6 94 80-90 4 98 90-100 2 100 100

(Majumdar, 2014)

Analysis:

It also calculates the average of the given data but it renders closed value to the central point. It get preferred over the mean due to its qualitative results and utilised as an base for decision making.

Media = 100/ 2 = 50

50 lie in the group of 30-40

= 30 + {(50-38)/16} * 10 = 37.50(Majumdar, 2014)

Median for the above table is 37.50

Mode:

Analysis: It is also another form of calculating average and in order to calculate the average the highest value in the frequency table is considered as mode. The value rendered by mode is considered as the benchmark and utilised in decision making for the purpose of comparison with others. In the given data mentioned as above the highest frequency is 20 and it is termed as mode for the data gathered(Majumdar, 2014).

## ii. Range and Standard Deviation. (AC 2.3)

Range: It is effective technique which get utilised in order to calculate maximum variations and it renders by measures of dispersion. In order to calculate the range lowest value gets deducted from the highest value. Below is the calculation made such as:

Range = (maximum value – minimum value)

= 20 – 2 = 18

18 customers become the range.(Majumdar, 2014)

Standard deviation:

 Amount Spend  (in £) No of orders Mid-value dx (x – 40 ÷ 10) F x dx Dx 10-20 18 15 -4 -72 288 20-30 20 25 -3 -60 180 30-40 16 35 -2 -32 64 40-50 14 45 -1 -14 14 50-60 12 55 0 0 0 60-70 8 65 1 8 8 70-80 6 75 2 12 24 80-90 4 85 3 12 36 90-100 2 95 4 8 32 100 -138 646

(Djokovic, 2013)

Standard deviation: {[646 / 138] – [138 *138 / 100 *100]} * 10

= {4.68 – 1.9044) *10

= 27.75

The standard deviation is £27.75 as there may be increase or decrease of £27.75 in the customer spending. This information is utilized for the decision making purpose.(Djokovic, 2013)

### iii. 25th Percentile (Lower Quartile) and 75th Percentile (Upper Quartile) and explain the use of Percentile (AC 2.4)

Percentile is a measure which get utilised under statistics where value below percentage is observed. Such as 25th percentile (lower quartile) are 3,5,6,8,9 in this 6 customers is lower quartile.

75th (upper quartile) are 12, 13, 16, 18, 21 in this 16 customers is the upper quartile.(Djokovic, 2013)

### iv. Inter-quartile range. (AC 2.4)

= Upper quartile – lower quartile

Upper quartile = 16

Lower quartile = 6

Inter- quartile range = 16 – 6 = 10 customers.(Zabukovec & Jaklic, 2015)

### v. Calculate correlation coefficient using the additional information provided below and discuss its advantages to a business. (AC 2.4)

Data given:

 Sales (units) Discount 20 1 40 4 50 6 55 6 60 10 70 12 80 13 90 14 100 15 565 81

Correlation with sales and discount is as discussed below such as:

 Sales (units) x Discount y xy x2 y2 x2y2 20 1 20 400 1 400 40 4 160 1600 16 25600 50 6 300 2500 36 90000 55 6 330 3025 36 108900 60 10 600 3600 100 360000 70 12 840 4900 144 705600 80 13 1040 6400 169 1081600 90 14 1260 8100 196 1587600 100 15 1500 10000 225 2250000 565 81 6050 40525 923 6209700

(Zabukovec & Jaklic, 2015)

X2 = 319,225

Y2 = 6,561

Correlation of coefficient = {n * ∑ xy – ∑x * ∑y) / SQRT (n * ∑x2 –(∑x) 2) * (n * ∑y2–(∑y) 2}

= [9 * 6050 – 565 * 81) / SQRT(9 * 40525 – 319225) * (9 * 923 – 6561)

= [54450 – 45765] / SQRT (45500) * (1746)

= 8685/ 8913.08

= 0.9744106(Zabukovec & Jaklic, 2015)

Correlation of coefficient = 0.9744106

Correlation of coefficient: It is termed as decision making tool and business management make use of it in their decision making purpose. It get utilised in order to build effective relationship among two variables and in the above calculation there is effective relationship is analysed among sales and discounts(Zabukovec & Jaklic, 2015).

Advantages of correlation of coefficient are:

• It is an effective technique which get utilised in order to make predictions related to the correlation.
• As if there is correlation among two variables then effective predictions get made for one variable by using second one.
• Effective relation is evaluating with the use of this technique.

(Zabukovec & Jaklic, 2015)

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