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2007
Thu
29
Mar
Miss Loi avatar

Histograms – Waiting Times for Cashier Aunty At Supermarket

(18)
Tuition given in the topic of E-Maths Tuition Questions from the desk of Miss Loi at 5:26 pm (Singapore time)

For the ‘开张典礼’ i.e. ‘opening ceremony’ of Miss Loi’s E-Maths Question of the Day section, here’s a very very very important question that will almost certainly appear in your exam (whatever your school is, wherever you are)!

This hails from the Sec 3 syllabus so Miss Loi reckons this will also be a timely memory refreshment exercise for those you who are now in Sec 4.

Due to complaints of customers snoring away while waiting to be served by the Cashier Aunty, a supermarket conducted a survey to find out the time spent by each of the 500 customers at its payment counter.

Time (t min)0 < t ≤ 11 < t ≤ 22 < t ≤ 33 < t ≤ 44 < t ≤ 6
No. of customers8313814010435
  1. Find the mean time spent by each customer at the Cashier Aunty counter.
  2. In a frequency distribution histogram drawn, the height of the column representing time t in the interval 4 < t ≤ 6 is 5 cm. Find the column height that represents t in 2 < t ≤ 3.

P.S. Please don’t blame Miss Loi if your own exam question doesn’t have any Cashier Aunty!

頑張って!!!

Revision Exercise

To show that you have understood what Miss Loi just taught you from the centre, you must:

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Comments & Reactions

18 Comments

  1. eastcoastlife's Avatar
    eastcoastlife commented in tuition class


    2007
    Mar
    30
    Fri
    5:10pm
     
    1

    Cool site!

    ok, I ask my son this question the next time we in the queue at the supermarket.

    Any questions related to MRT, McDonald's, Crystal Jade, Toto (oops, paiseh, tats for Daddy!) - welcome! Must keep testing ah boi in case his brain gets too much rest. hehe.....

  2. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Mar
    31
    Sat
    9:10am
     
    2

    ECL: on behalf of the Cashier Aunty, Miss Loi would like to say 欢迎光临!!!

  3. kiroii's Avatar
    kiroii commented in tuition class


    2007
    Sep
    5
    Wed
    8:10pm
     
    3

    1) 2.282

    2) uh frequency distribution histogram? Isit synomous to frequency density? If thats de case then shouldnt it be class frequency / class width? If it is then..is the answer 138?

  4. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Sep
    8
    Sat
    2:07pm
     
    4

    Oops kiroii,

    1) Your answer to the mean is a little bit off. Note that this is a set of grouped data with ranges. So you'll need to use a value of t in the middle of each range.

    You can find the mean via two ways. The first way would be your usual group data formula {Sigma ft}/{Sigma f}.

    If time permits in your exam, you can also use the assumed mean formula a + {Sigma fd}/{Sigma f} (where a is your own assumed mean and d is the deviation (t - a) from the assumed mean) to check your answer.

    They should both yield the same answer. Try it and show Miss Loi your workings!

    P.S. Oh BTW it's always a good practice to show all your calculated values like ft and fd and their corresponding ranges in table form to reduce your chances of careless mistakes!

  5. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Sep
    8
    Sat
    2:56pm
     
    5

    2) Wrong wrong wrong!!!

    Take note that there are unequal class widths in this histogram, so you're need to first find the proportionality factor k for this particular histogram.

    If you recall from your textbook,

    (height of rectangle x class width) α class frequency

    Hence,

    height of rectangle α (class frequency/class width)

    which can also be written as

    height of rectangle = k x class frequency/class width

    Note α means proportional not equal, so you can't simply just divide like you did! You'll need to find k first.

    With all these hints can you solve this now???

  6. kiroii's Avatar
    kiroii commented in tuition class


    2007
    Sep
    8
    Sat
    3:21pm
     
    6

    er is de ans for 2) 38.33?

    5= k(36/2) [5 is de height, 36/2 is classfrequency/ class width]

    k= 10/36

    therefore h = k x (138/1)
    h = 38.33

  7. kiroii's Avatar
    kiroii commented in tuition class


    2007
    Sep
    8
    Sat
    3:25pm
     
    7

    eh for question 1

    [(83 x 0.5)+(138 x 1.5)+(140x2.5)+(3.5x104)+(5x35)] / (83+138+140+104+35)

    i did use de middle number

    lo and behold after simplication..
    2.275

  8. kiroii's Avatar
    kiroii commented in tuition class


    2007
    Sep
    8
    Sat
    3:26pm
     
    8

    eh ps: i have no idea how did i get 2.285 probably typo error>.

  9. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Sep
    8
    Sat
    7:22pm
     
    9

    1) 2.275 is correct and your working's fine too. But you're really prone to typos - first you said 2.282 and now 2.285!

    2) Careless careless careless once more! Why is your frequency 36? And under which range are you supposed to find the height?

    Slow down and look at the table first!

  10. kiroii's Avatar
    kiroii commented in tuition class


    2007
    Sep
    9
    Sun
    1:24am
     
    10

    oh..zz mistook 35 for 36..how inappropriate..
    anyway here's my revised workings

    5= k(35/2) [5 is de height, 36/2 is classfrequency/ class width]

    k= 10/35

    therefore h = k x (138/1)
    h = (10/35) x (138)
    h = 39.4

  11. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Sep
    9
    Sun
    4:24pm
  12. Noel's Avatar
    Noel commented in tuition class


    2007
    Oct
    29
    Mon
    5:00pm
     
    12

    erm can i do it this way?
    frequency density of 4

  13. ME's Avatar
    ME commented in tuition class


    2007
    Oct
    30
    Tue
    12:56am
     
    13

    hah. lazy type the working.

    i got 2.275 for the first answer.

    and

    40cm for the column height.

    (:

  14. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Oct
    30
    Tue
    1:12am
     
    14

    Yes

    h = k x (140/1)
    h = (10/35) x (140) = 40cm

    Please LOOK at the frequency table or histogram properly and don't make the same kind of careless mistakes listed here!

    Ok passed. How come so late still here? Tomorrow morning's Paper 2 postponed is it???

  15. ME's Avatar
    ME commented in tuition class


    2007
    Oct
    30
    Tue
    6:36pm
     
    15

    haha.. it should be today. (:
    and it's not late. it was in the morning. =D

    anyway, i screwed it up.. XD..

  16. fransiska's Avatar
    fransiska commented in tuition class


    2008
    Oct
    28
    Tue
    12:46am
     
    16

    ms loi,
    is there a formula for histogram???for qs 2?how come i dont understand???its scaring meeeee....

  17. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2008
    Oct
    29
    Wed
    12:20am
     
    17

    Alright Fransiska let's see if this diagram helps you:

    Cashier Aunty Histogram Columns

    In part 2 of the question, we're trying to find the column height (the ? in the diagram) for 2 < t ≤ 3. At the same time, we're told that the column height for 4 < t ≤ 6 is 5cm.

    Now the important thing to note (esp. for histograms with unequal intervals like this one) is that the class frequency is represented by the area of the column (=*'class height' x class width). So the 'formula' here is:

    *'Class height' of column = class frequency/class width
    ⇒ Actual height of column = k x class frequency/class width

    * This 'class height' is actually the frequency density.

    We have to include a k here as the actual height is only proportional to the frequency density (they're typically NOT equal):
    actual height α frequency density ⇒ actual height = k x frequency density.

    Since its given that the actual column height for 4 < t ≤ 6 is 5cm, we need to know the value of k first to find the column height for 2 < t ≤ 3.

    ∴ 5cm = k * 35/2 doubleright k = 10/35

    With this value of k, we can then find the column height for 2 < t ≤ 3

    ∴ Height = (10/35)*{140/1} = 40cm

    QUICK METHOD:

    Since Noel mentioned frequency density earlier, you can actually obtain the answer quickly if you're familiar with the concept i.e. frequency density = class frequency/class width,

    Freq. density for 4 < t ≤ 6 = 35/2 = 17.5 → represented by height of 5cm
    Freq. density for 2 < t ≤ 3 = 140/1 = 140 → let this height be x cm

    So comparing the ratios,
    x/5 = 140/17.5x = 40cm 😉

  18. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2008
    Oct
    29
    Wed
    10:14pm
     
    18

    Just for the sake of completing the picture, Miss Loi shall proceed to plot the histogram for this set of data.

    Whenever you're asked to plot a histogram, always remember to 'standardize' the class frequencies to frequency densities before plotting, especially when you have unequal intervals!

    Time (t min)Class widthClass frequencyFrequency Density
    0 < t ≤ 118383÷1=83
    1 < t ≤ 21138138÷1=138
    2 < t ≤ 31140140÷1=140
    3 < t ≤ 41104104÷1=104
    4 < t ≤ 623535÷2=17.5

    So the histogram will look like this (note the 'compressed' red column):

    Cashier Auntie Histogram

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