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Solve theis Math questions - Speech or Presentation Example

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The correlation between variables is 0 that shows non-linear relationship. However, polynomial curve captures all the data points on set A that shows variables…

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EBS101: Statistical Methods Assignment Sheet 2 - Submit answers by 4pm on 27 February 2009 For with a record number ending in 2 Pearson product moment correlation coefficient
Correlation
EXCEL functions
SET A
0.0000
CORREL or PEARSON
SET B
0.9589
CORREL or PEARSON
SET C
-0.9912
CORREL or PEARSON
Regression line for set A data set do not capture any data points, therefore shows no association between variables. The correlation between variables is 0 that shows non-linear relationship. However, polynomial curve captures all the data points on set A that shows variables have perfect association (polynomial) between them.
Regression line for set B data set captures approximately all data points, therefore shows strong association between variables. The correlation between variables is 0.96 that shows strong positive linear relationship.
Regression line for set C data set captures all data points, therefore shows strong association between variables. The correlation between variables is 0.99 that shows strong (or perfect) negative linear relationship.
Item
Fuel/gallon
Oil/litre
Tyres
Insurance
Cost of Motoring Index
Quantity (per year) ()
1000
15
2
1
Base cost ()
1.30
1.50
80.00
300.00
Cost in year 1
1.12
2.00
130.00
650.00
115.57
Cost in year 2
1.16
2.00
145.00
575.00
115.29
Cost in year 3
1.18
2.00
175.00
550.00
118.37
Cost in year 4
1.20
2.25
190.00
550.00
121.39
Cost in year 5
1.25
1.80
200.00
575.00
126.34
1
2.5
3.00
1.25
2
2.9
3.26
1.36
3
3.0
3.32
1.39
4
3.2
3.44
1.44
5
4.3
4.02
1.67
6
4.8
4.27
1.76
7
6.4
4.96
2.00
8
7.8
5.50
2.17
9
9.3
6.02
2.33
10
11.7
6.77
2.54
Mean
5.59
4.45
1.79
SD
3.13
1.30
0.45
The transformation has reduced the value of mean and standard deviation. Both are non-linear transformation that leads to reduced skewness of distribution of data.
Probability
EXCEL functions
Pr{Z > −1.96}
0.9750
1-NORMSDIST(-1.96)
Pr{−1.64 < Z < 1.96}
0.9245
NORMSDIST(1.96)-NORMSDIST(-1.64)
Pr{Z < 1.96}
0.9750
NORMSDIST(1.96)
Pr{Z > −1}
0.8413
1-NORMSDIST(-1)
Pr{Z < −1 or Z > 1.64}
0.2092
NORMSDIST(-1)+(1-NORMSDIST(1.64))
Pr{−1 < Z < 1.64}
0.7908
NORMSDIST(1.64)-NORMSDIST(-1)
Pr{Z < 1}
0.8413
NORMSDIST(1)
Pr{−1 < Z < 1}
0.6827
NORMSDIST(1)-NORMSDIST(-1)
Pr{0 < Z < 1.64}
0.4495
NORMSDIST(1.64)-NORMSDIST(0)
Pr{−1 < Z < 1.96}
0.8163
NORMSDIST(1.96)-NORMSDIST(-1)
EXCEL functions
Pr{Z > } = 0.05
= 1.64
NORMSINV(1-0.05)
Pr{Z < } = 0.1
= -1.28
NORMSINV(0.1)
Pr{− < Z < } = 0.95
= 1.96
NORMSINV((1-0.95)/2)
Pr{0 < Z < } = 0.1
= 0.25
NORMSINV(0.5+0.1)
Pr{|Z| < } = 0.5
= 0.67
NORMSINV((1-0.5)/2)
Standardisation (Z)
Probability
EXCEL functions
Pr{X < 40}
-2
0.0228
NORMSDIST(-2)
Pr{X < 70}
1
0.8413
NORMSDIST(1)
Pr{50 < X < 80}
2
-1
0.8186
NORMSDIST(2)-NORMSDIST(-1)
Pr{X > 75}
1.5
0.0668
1-NORMSDIST(1.5)
Pr{X < 45 or X > 80}
2
-1.5
0.0896
(1-NORMSDIST(2))+NORMSDIST(-1.5)
Or using EXCEL function NORMDIST(x, mean, standard_dev, cumulative)
Probability
EXCEL functions
Pr{X < 40}
0.0228
NORMDIST(40,60,10,TRUE)
Pr{X < 70}
0.8413
NORMDIST(70,60,10,TRUE)
Pr{50 < X < 80}
0.8186
NORMDIST(80,60,10,TRUE)-NORMDIST(50,60,10,TRUE)
Pr{X > 75}
0.0668
1-NORMDIST(75,60,10,TRUE)
Pr{X < 45 or X > 80}
0.0896
NORMDIST(45,60,10,TRUE)+(1-NORMDIST(80,60,10,TRUE))
Since, µ = 60 and σ =10, therefore any negative value of X will be more than 6 standard deviation from the mean and the probability below that will be zero.
For example: Pr{X < -10}
Pr{X < -10} = 0.00000000
Using Excel function NORMINV(probability, mean, standard_dev)
EXCEL functions
Pr{X > } = 0.05
= 76.45
NORMINV((1-0.05),60,10)
Pr{X < } = 0.1
= 47.18
NORMINV(0.1,60,10)
Pr{−< X < } = 0.95
= 76.45
NORMINV((0.95-0),60,10)
Pr{μ < X < } = 0.1
= 62.53
NORMINV((0.5+0.1),60,10)
Pr{|X| < } = 0.5
= 60.00
NORMINV((0.5-0),60,10)
µ = 500 and σ =16, and
What proportion of bags weigh less than 516 grammes?
Using Excel function NORMDIST area below Z = 1 is 0.8413.
Therefore, 0.8413 (or 84.13%) of bags will weigh less than 516 grammes.
What proportion of bags weigh between 476 and 516 grammes?
Using Excel function NORMDIST area below Z = 1 and Z = -1.5 are 0.8413 and 0.0668.
Area between Z equal to -1.5 to 1 = 0.8413 – 0.0668 = 0.7745
Therefore, 0.7745 (or 77.45%) of bags will weigh between 476 and 516 grammes.
If the manufacturer wants to keep the mean weight at 500 grammes, but wishes to adjust the production process so that 2.5% of bags weigh less than 495 grammes, calculate the value the standard deviation should take.
Using Excel function NORMSINV the value of Z for 0.025 (or 2.5%) is -1.96.
The value the standard deviation should take will be approximately 2.55.
Paperback: = 240, and = 12
Hardbacks: = 300, and = 16
a) The probability that one paperback and one hardback together weigh less than 570 grammes, and
Combined weight = 570 grammes
Combined mean, µ= + = 240 + 300 = 540
Standard deviation, σ = =
Using Excel function NORMDIST area below Z = 1.5 is 0.9332.
Hence, Pr{combined weight < 570 gm} = 0.9332
There is 93.32 percent chance that one paperback and one hardback together weigh less than 570 grammes.
b) The probability that a selection of 9 different hardbacks weighs between 2676 and 2772 grammes, and
Combined mean of 9 hardbacks = = 2700 grammes
Standard deviation, σ = = 48
Using Excel function NORMDIST area below Z = -0.5 and Z = 1.5 are 0.3085 and 0.9332.
Hence, Pr{2676 < 9 different hardbacks < 2772 gm} = 0.9332 – 0.3085 = 0.6247
There is 62.47 percent chance that a selection of 9 different hardbacks weighs between 2676 and 2772 grammes.
c) 95% limits for the weight of a random selection of 4 different paperbacks
The bookshop also supplies other bookshops, which usually order multiple copies of the same book.
Using Excel function NORMSINV, Z value for 95% limit is 1.645
Combined mean weight of 4 paperbacks = 4 = = 960
Standard deviation, σ = = 24

= 999.48 grammes.
95% limits for the weight of a random selection of 4 different paperbacks is 999.48 grammes.
d) One such order is for 4 copies of a particular paperback. Find 95% limits on the weights of such orders, and explain why are these different from the 95% limits in (c).
Using Excel function NORMSINV, Z value for 95% limit is 1.645
Average weight of 4 particular paperbacks = = 240
Standard deviation, σ =

= 249.87 grammes.
95% limits for the average weight of 4 copies of a particular paperback is 249.87 grammes.
These are different from part c because here 95% limits on average weight for 4 particular order is asked whereas in part c, 95% limits for the weight of a random selection of 4 different paperbacks is asked.
µ = 3000, and σ = 300, cost of stand = £3000, other cost = £2700
p (order) = 0.05, profit on each order = £50
The probability that attending the fair generates enough orders to cover your costs,
Total cost = cost of stand + other cost = £3000 + £2700 = £5700
Number of order required to cover total cost = 5700/50 = 114
Therefore, number of enquiry = 114/0.05 = 2280
Using Excel function NORMDIST area below Z = -2.4 is 0.0082.
Hence, Pr{Enquiries > 2280} = 1 – 0.0082 = 0.9918
There is 99.18 percent chance that attending the fair generates enough orders to cover my costs.
The probability that attending the fair leads to a profit of more than £2150
Profit + Total cost = £2150 + £5700 = £7850
Number of order required to cover profit = 7850/50 = 157
Therefore, number of enquiry = 157/0.05 = 3140
Using Excel function NORMDIST area below Z = 0.4667 is 0.6796.
Hence, Pr{Enquiries > 3140} = 1 – 0.6796 = 0.3204
There is 32.04 percent chance that attending the fair leads to a profit of more than £2150. Read More
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