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 |

- Find the mean time spent by each customer at the Cashier Aunty counter.
- 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!

## 18 Comments

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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.....

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

曜

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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?

曜

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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 oftin the middleof 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 meanformula (whereais your own assumed mean anddis 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

ftandfdand their corresponding ranges in table form to reduce your chances of careless mistakes!曜

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2) Wrong wrong wrong!!!

Take note that there are

unequalclass widths in this histogram, so you're need to first find theproportionality factorfor this particular histogram.kIf 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 =

kx class frequency/class widthNote α means proportional

notequal, so you can't simply just divide like you did! You'll need to findkfirst.With all these hints can you solve this now???

曜

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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

曜

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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

曜

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

曜

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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!

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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

曜

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The range! The range!!!

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erm can i do it this way?

frequency density of 4

曜

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hah. lazy type the working.

i got 2.275 for the first answer.

and

40cm for the column height.

(:

曜

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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???

曜

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haha.. it should be today. (:

and it's not late. it was in the morning. =D

anyway, i screwed it up.. XD..

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ms loi,

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

曜

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Alright

Fransiskalet'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

unequalintervals like this one) is that the class frequency is represented by theareaof the column (=*'class height' x class width). So the 'formula' here is:*'Class height' of column = class frequency/class width

⇒ Actual height of column =

kx class frequency/class width* This 'class height' is actually the

frequency density.We have to include a

khere as the actual height is only proportional to the frequency density (they're typically NOT equal):actual height α frequency density ⇒ actual height =

kx frequency density.Since its given that the actual column height for 4 <

t≤ 6 is 5cm, we need to know the value ofkfirst 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

Noelmentionedfrequency densityearlier, 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 5cmFreq. density for 2 <

t≤ 3 = = 140 → let this height bexcmSo comparing the ratios,

⇒

x= 40cm 😉曜

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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 densitiesbefore plotting, especially when you have unequal intervals!tmin)t≤ 1t≤ 2t≤ 3t≤ 4t≤ 6So the histogram will look like this (note the 'compressed' red column):