Question

Need help :C

A bakery in Hong Kong regularly orders carton of cranberries from Australia. The net weight (in ounces) of

cranberries of a random sample of 20 cartons were as follow:

160 170 170 190 210 240 240 240 270 280

290 290 300 420 500 520 530 610 630 650



(c) The bakery orders the cranberries from Australia with currency exchange rate of AUD 1 to HKD 5.8.

For each ounce of Australia cranberries costs AUD 2 and the fixed shipping fee of each carton is HKD

80. Find the mean, standard deviation, range, and third quartile of the payment (in HKD) of ordering a

carton of cranberries from Australia to Hong Kong.

Answer 1

i) Mean = 1933.1817

ii) Range = 5684

iii) Third quartile = 6054

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{((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