Unit 2 Business Decision Making Assignment Help

Unit 2 Business Decision Making Assignment Help

This is a solution of Unit 2 Business Decision Making Assignment Help that describes about Developing business

Task 2

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 25thpercentile (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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