MATHS (SECTION A)

Question

SECTION  A

The following table shows the distribution of the numbers of hours talking on mobile phones by a group of students on a certain day. Number of hours talking on mobile phone1234Number of students8ab12 It is given that a and b are positive numbers. (a) Find the least possible value and the greatest possible value of the median of the distribution. (b) If b =2 and the inter-quartile range of the distribution is 3, how many possible values of a are there? Explain your answer.
(5 marks)

wy · Accepted Answer

For the IRQ to be 3, the upper quartile must be 4 and the lower quartile must be 1.
Total no. of students = 8 + a + 2 + 12 = 22 + a
so no. of students in each quartile = (22 + a)/4.
For the lower quartile to be in the 1 hour zone, 
(22 + a)/4 < 8
22 + a < 32
a < 10
Since 0 is not a positive integer, so possible value of a is 1,2,.... 9. Total 9 possible values.

Stardust · Answer

Thanks for your teaching.^^

[**It is given that a and b are positive numbers.

So there are 9 cases.]

2014-01-08 18:35:04 補充：
你寫得好清楚,好易明@@b

[我老師叫我地諗晒所有cases,
我心諗(b) 得3分, 仲要section a, 邊洗咁麻煩- -]

知足常樂 · Answer

前幾天有人問過這題～

2014-01-08 18:05:18 補充：
OK, 咁我只講 part (b) 就可以了。

我覺得最簡單的方法係咁諗：

IQR = 3 咁一定係 UQ = 4，LQ = 1

先知道 a = 0 是可以的。

現在你想知道 a 最大可以是多少，那麼 加１就是答案。

2014-01-08 18:08:33 補充：
１的數目比４少。

所以你要保持１是LQ，比保持４是UQ難。

且看看加多少個２仍能使１是LQ。

１本身有８個，如果１仍要是LQ，你最多只能把７個放在以下。

那麼a最大的情況只能是：
{7個1} {LQ = 1} {7個} {median} {7個} {UQ = 4} {7個}

所以總數 = 7 + 1 + 7 + 1 + 7 + 1 + 7 = 31

所以 max a = 31 - 8 - 2 - 12 = 9

所以 a 可以由 0 至 9, 共 10 個cases

2014-01-08 18:22:10 補充：
是啊～

a ≠ 0～

好～

那 a can be either of {1, 2, 3, 4, 5, 6, 7, 8, 9}

2014-01-08 18:25:34 補充：
阿月，係呢題：
http://hk.knowledge.yahoo.com/question/question?qid=7013122600075

內容唔同，但數字完全一樣，唔知係咩題目～
可能係抄書～

MATHS (SECTION A)

3 個解答