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# Histograms – Waiting Times for Cashier Aunty At Supermarket

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

Updated on

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 ≤ 1 1 < t ≤ 2 2 < t ≤ 3 3 < t ≤ 4 4 < t ≤ 6 No. of customers 83 138 140 104 35
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!

### Revision Exercise

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

1. 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 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 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 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 .

If time permits in your exam, you can also use the assumed mean formula (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 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 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 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 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 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 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 Friend Miss Loi on Facebook @MissLoi commented in tuition class

2007
Sep
9
Sun
4:24pm
12. 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 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 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 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 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 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:

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 =

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

∴ Height = = 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 = = 17.5 → represented by height of 5cm
Freq. density for 2 < t ≤ 3 = = 140 → let this height be x cm

So comparing the ratios,
x = 40cm 😉

18. 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):

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