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During the course of an hour, an employee at Ultimate...

This topic has 2 expert replies and 0 member replies

During the course of an hour, an employee at Ultimate...

Post Fri Nov 10, 2017 9:42 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    During the course of an hour, an employee at Ultimate Packing Solutions, wrapped packages weighing 48, 32, 63, 12, 40 and 8 pounds. What was the median weight of the packages that the employee packed?

    A. 33
    B. 33.333
    C. 34
    D. 36
    E. 62

    The OA is D.

    Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.

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    Post Fri Nov 10, 2017 11:20 am
    Hi swerve,

    When determining the MEDIAN of a group of numbers, we first have to put the numbers in order (from least to greatest), then find the 'middle number.' When there is an EVEN number of terms, then the MEDIAN is the AVERAGE of the two 'middle terms.'

    Here, we have 6 numbers:
    8, 12, 32, 40, 48, 63

    The MEDIAN of this group is the average of 32 and 40... 72/2 = 36.

    Final Answer: D

    GMAT assassins aren't born, they're made,
    Rich

    _________________
    Contact Rich at Rich.C@empowergmat.com

    Thanked by: swerve
    Post Fri Nov 10, 2017 12:35 pm
    Quote:
    During the course of an hour, an employee at Ultimate Packing Solutions, wrapped packages weighing 48, 32, 63, 12, 40 and 8 pounds. What was the median weight of the packages that the employee packed?

    A. 33
    B. 33.333
    C. 34
    D. 36
    E. 62

    The OA is D.

    Please, can any expert explain this PS question for me? I have many difficulties to understand why that is the correct answer. Thanks.
    Hi Swerve,
    Let's take a look at your question.

    We are asked to find the median weight of the packages weighing 48, 32, 63, 12, 40, 8.
    To find median of ungrouped data, we arrange the data in ascending or descending order.

    The weights in ascending order can be written as:
    $$8,\ 12,\ 32,\ 40,\ 48,\ 63$$

    When the number of values is odd, the median is equal to the middle value of the data arranged in ascending or descending order but When the number of values is even the median is the average of the two middle values of the data arranged in ascending or descending order.

    In this data, we can see that we have six values.
    The middle two values are 32 and 40.
    Therefore,
    $$Median\ =\ \frac{32+40}{2}$$
    $$Median\ =\ \frac{72}{2}=36$$

    Therefore, Option D is correct.

    Hope it helps.
    I am available if you'd like any help.

    _________________
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    Thanked by: swerve
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