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3 years ago

Need help :C

Mathematics

1 answer:

rjkz [21]3 years ago

7 0

Answer:

i) Mean = 1933.1817

ii) Range = 5684

iii) Third quartile = 6054

Step-by-step explanation:

Given data :

currency exchange rate : 1 AUD = 5.8 HKD

cost of each ounce = 2 AUD

Fixed shipping cost for each carton = 80 HKD

number of cartons = 20

next determine the total cost of the 20 cartons in HKD

= ∑(weight in ounce * cost of each ounce *exchange rate) +fixed shipping cost

= ∑ ( 160*2*5.8 + 80 ) + -------------- + (650 *2*5.8 + 80 ) ----------------- ( 1 )

= 81756 HKD

i) find the mean value ( X )

= Total cost / number of cartons

= 81756 / 20 = 4087.8

ii) Find the standard deviation

= $\sqrt{\frac{(1936-4087.8)^2+-----+(7620-4087.8)^2}{20-1} }$ note: std = √∑(xi-X )^2 / (n-1)

= 1933.1817

iii) Find the range

Range = highest cost - lowest cost ( values gotten from equation 1 )

= 7650 - 1936

= 5684

iv) Determine the third quartile

third quartile = 6054

attached below is the detailed solution

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