A specialty tea shop sells five distinct sizes of loose-leaf tea canisters. The table belo… — GRE General Test

A specialty tea shop sells five distinct sizes of loose-leaf tea canisters. The table below shows the weight per canister and the number of canisters sold during a promotional week:

Canister Size	Weight per Canister (ounces)	Number of Canisters Sold
Sampler	5	16
Small	12	18
Medium	15	$n$
Large	20	6
Executive	25	4

If the mean weight of all canisters sold during the week is $13$ ounces, what is the median weight, in ounces, of the canisters sold?

A
$12$ ounces
B
$13$ ounces
$15$ ouncesAnswer
D
$15.4$ ounces
E
$16$ ounces

Answer

15 ounces

The correct answer is 15 ounces. Calculating the total weight as a function of

n

gives

516 + 15n

, and the total number of canisters as

44 + n

. Setting the mean

\frac{516 + 15n}{44 + n} = 13

yields

n = 28

. The total number of canisters sold is

44 + 28 = 72

. For 72 items sorted in ascending order, the median is the average of the 36th and 37th values. Cumulative frequencies show that the 1st through 16th items are 5 ounces, the 17th through 34th items are 12 ounces, and the 35th through 62nd items are 15 ounces. Thus, both the 36th and 37th items weigh 15 ounces, making the median 15 ounces.

Step-by-Step Solution

Set up an equation for the total number of canisters and total weight.

Total canisters =

16 + 18 + n + 6 + 4 = 44 + n

. Total weight =

5(16) + 12(18) + 15(n) + 20(6) + 25(4) = 516 + 15n

ounces.

The mean of a weighted dataset is the total sum of all values divided by the total number of items.

Solve for the missing frequency

n

using the given mean of 13 ounces.

\frac{516 + 15n}{44 + n} = 13 \implies 516 + 15n = 13(44 + n) \implies 516 + 15n = 572 + 13n \implies 2n = 56 \implies n = 28

Equating the weighted average expression to 13 allows us to determine the exact number of Medium canisters sold.

Determine the total number of items and locate the position of the median.

Total canisters =

44 + 28 = 72

. The median is the average of the

36^{\text{th}}

and

37^{\text{th}}

values when ordered.

For an even number of data points

N = 72

, the median is the arithmetic mean of the

(N/2)^{\text{th}}

and

(N/2 + 1)^{\text{th}}

items.

Compute the cumulative frequency distribution to find the values at the

36^{\text{th}}

and

37^{\text{th}}

positions.

5 oz: positions 1 to 16; 12 oz: positions 17 to 34; 15 oz: positions 35 to 62. Both

36^{\text{th}}

and

37^{\text{th}}

positions have a weight of 15 ounces, so the median is 15 ounces.

Cumulative counting reveals that items in positions 35 through 62 all correspond to 15-ounce canisters.

Key Concept

Weighted mean and finding the median of frequency distribution data

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