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Anna71 [15]
3 years ago
10

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

<u>i) find the mean value ( X ) </u>

= 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

<u>iii) Find the range </u>

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

           = 7650 - 1936

           = 5684

<u>iv) Determine the third quartile </u>

third quartile = 6054

<em>attached below is the detailed solution</em>

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