stats question

[muzirox]

Hi guide in the following question plzzzzzzzzzz,this is from 2003-p1-q11(c) :

Bag X contains 8 blue counters and 5 red counters.
Bag Y contains 9 blue counters and 6 red counters.

A counter is selected at random from Bag X and placed in Bag Y. A second counter is then
selected at random from Bag Y and then placed in Bag X.

Calculate the probability that the number of blue counters and red counters in Bag X and in
Bag Y remains unchanged after the two selections. [6]

 

[Alchemist4lyf]

Alright. Thing is this. We want the end product to remain the same, that is, we want 8b 5r in X, and 9B, 6R in Y by the end. So, what we'll have to do, is select either the Blue, take it to Y, then place it back. Or either the Red, take it to Y, then place it back.

So, let's consider the possibilities. X -8B, 5R. Y- 9B, 6R.
The probability would be, P(getting same situation after the two selections) = P(Taking a blue from X, Taking a blue from NEW Y) + P(Taking a red from X, Taking blue from NEW Y)

= P(B x B) + P(R x R)
So P (B from X) = 8/ 13
P (B from New Y) = 10/16 (Because a blue went there, so Blue counters increased, and so did the total).
P (R from X) = 5/ 13
P (R from New Y) = 7/16 (Because a red wen there, so Red counters increased, and so did the total).
So we have P(Unchanged) = (8/13 x 10/16) + (5/13 x 7/16)
= 115/208. Which happens to be the correct answer.
Hope i made it seem easy for you.

 

[muzirox]

Alright. Thing is this. We want the end product to remain the same, that is, we want 8b 5r in X, and 9B, 6R in Y by the end. So, what we'll have to do, is select either the Blue, take it to Y, then place it back. Or either the Red, take it to Y, then place it back.

So, let's consider the possibilities. X -8B, 5R. Y- 9B, 6R.
The probability would be, P(getting same situation after the two selections) = P(Taking a blue from X, Taking a blue from NEW Y) + P(Taking a red from X, Taking blue from NEW Y)

= P(B x B) + P(R x R)
So P (B from X) = 8/ 13
P (B from New Y) = 10/16 (Because a blue went there, so Blue counters increased, and so did the total).
P (R from X) = 5/ 13
P (R from New Y) = 7/16 (Because a red wen there, so Red counters increased, and so did the total).
So we have P(Unchanged) = (8/13 x 10/16) + (5/13 x 7/16)
= 115/208. Which happens to be the correct answer.
Hope i made it seem easy for you.

thanx a lot boi u've solved my prob

 

[muzirox]

One more question plzz
Qs from-2007 p2 (3)

The values of a variable, X, are formed into a grouped frequency distribution, with classes stated
as 0 – 4, 5 – 9, 10 – 14, etc.
For the 5 – 9 class, give the true class limits and the class mid-point in each of the following
cases.
(i) x represents the number of cars parked in a car park at noon on each day during one month.
[2]
(ii) x represents the length, rounded to the nearest millimetre, of the leaves on a particular plant.
[2]
(iii) x represents the number of completed kilometres run in a fixed amount of time by each of a
number of athletes. [2]

plzzzzzz solve this

 

[Alchemist4lyf]

5 - 9 Class.

(i). Lower Class Limit = 5. Upper Class Limit = 9. (I'm not sure about this one, really.)
(ii). Lower Class Limit = 5 - (0.5) = 4.5 Upper Class Limit = 9 + (0.5) = 9.5. (I'm sure about this one, because in such cases, where the continuous data seems to be discrete, you have to make it continous by seeing how much it is rounded by. The Formula for it is = Lower Class Limit = Rounded Lower Class Limit - 0.5R. Upper Class Limit = Rounded Upper Class Limit + 0.5R. R is how much it has been rounded by. In this case it is 1mm, so L.C.L = 5 - 0.5(1). U.C.L = 9 + 0.5(1).
(iii). Lower Class Limit = 5. Upper Class Limit = 10. I'm sure about this one too, because in this case, the answer can't be in points, because that would not be a COMPLETED RUN. So what you do it, Lower Class Limit of First Class - Lower Class Limit of Second Class.

I hope i helped you out. I really don't know about the first one, because I do these type of questions very carelessly.

 

[shahzaibzindani]

(i). Lower Class Limit = 5. Upper Class Limit = 9. (because no. of cars cannot be in decimals so as it is a discreet data the true class limits will remain the same)
(ii). Lower Class Limit = 4.5. Upper Class Limit = 9.5. (it is a continous data, the gap between the lower class boundary and the upper class boundary i.e 1 is divided by 2, i.e 0.5 so it is to be subtracted from lower class boundary i.e (5-0.5 = 4.5) and added to upper class boundary i.e (9+0.5= 9.5)
(iii). Lower Class Limit = 5. Upper Class Limit = 10 (because in this case, the answer can't be in points, because that would not be a COMPLETED RUN)
Hope it is helpful

 

[muzirox]

(i). Lower Class Limit = 5. Upper Class Limit = 9. (because no. of cars cannot be in decimals so as it is a discreet data the true class limits will remain the same)
(ii). Lower Class Limit = 4.5. Upper Class Limit = 9.5. (it is a continous data, the gap between the lower class boundary and the upper class boundary i.e 1 is divided by 2, i.e 0.5 so it is to be subtracted from lower class boundary i.e (5-0.5 = 4.5) and added to upper class boundary i.e (9+0.5= 9.5)
(iii). Lower Class Limit = 5. Upper Class Limit = 10 (because in this case, the answer can't be in points, because that would not be a COMPLETED RUN)
Hope it is helpful

thats correct bcuz recently mu teacher has told me that,thanx for the answer

 

[muzirox]

New qs:O/N 2006 p2 (1)
these qs r really hard for me

(a) A white dice and a blue dice are thrown. Each dice is unbiased and has faces numbered 1, 2,
3, 4, 5 and 6.
Events A, B, C and D are defined below.
A: 3 is scored on the white dice.
B: 3 is scored on the blue dice.
C: 3 is scored on both dice.
D: 4 is scored on both dice.
(i) Name two of these events which are mutually exclusive. [1]
(ii) Name two of these events which are independent. [1]

(b) Given that events E and F are independent, state, for each of the following, whether or not it
must be true.
(i) P(E) = P(F). [1]
(ii) P(E and F) = P(E) + P(F). [1]
(iii) P(E and F) = P(E) × P(F). [1]
(iv) P(E and F) = 0. [1]

 

[Alchemist4lyf]

(a)(i). Events that are mutually exclusive are C and D. This is because if one of them occurs, the other can not. For instance, if we get a (3,3). Then, we can not get a (4,4).
(ii). Events that are independent are A and B. This is because, scoring a 3 on the white dice, will do nothing to the probability of scoring a 3 on the blue dice, i.e. they are independent.
(b)(i). Not true because independent does not mean that their probabilities are necessarily same. It just means that their probabilities are independent of one another.
(ii).Not true because independent events have the formula P(E and F) = P(E) x P(F).
(iii). True.
(iv). Not true because this equation applies to mutually exclusive events, where the probabilities of both events occurring are 0.

Hope I helped. I remember doing this question quite carelessly too. I haven't check the Examiner Report. If I'm wrong in any part, please do tell me.

 

[muzirox]

(a)(i). Events that are mutually exclusive are C and D. This is because if one of them occurs, the other can not. For instance, if we get a (3,3). Then, we can not get a (4,4).
(ii). Events that are independent are A and B. This is because, scoring a 3 on the white dice, will do nothing to the probability of scoring a 3 on the blue dice, i.e. they are independent.
(b)(i). Not true because independent does not mean that their probabilities are necessarily same. It just means that their probabilities are independent of one another.
(ii).Not true because independent events have the formula P(E and F) = P(E) x P(F).
(iii). True.
(iv). Not true because this equation applies to mutually exclusive events, where the probabilities of both events occurring are 0.

Hope I helped. I remember doing this question quite carelessly too. I haven't check the Examiner Report. If I'm wrong in any part, please do tell me.

thanx once again,but the prob is why can't A & B be mutually exclusive ,C & D be independent,plzzz explain this as i have prob in judging btween them,waiting for ur comprehensive reply

 

[Alchemist4lyf]

Mutually exlusive events are like this. Suppose you have A and B. A is heads, B is tails. Now if you get a heads, the probability of tails is 0, and vice versa. So over here, A is a three on WHITE DICE. And B is a three on BLUE DICE. So if you get, for suppose, a three on the white dice, the probability of a three on blue dice is NOT 0, because the two dice are not related.
Consider this. A is a 3 on Blue Dice, and B is a 4 on Blue Dice. Now this is mutually exclusive, because only one can occur, that is, you can not get both. But in the question, you can get both, a 3 on Blue and a 3 on White, because there are two die. So it is NOT mutually exclusive. Hope that helped out.

Now, independent events are like this. If there are two events, they will be independent only if the probability of one of them occurring, does not affect the probability of the second one of them occurring. So, if you look over here, C and D can not be independent. This is because, the probability of a three on both dice, will reduce the probability of 4 on both dice. Over here, we are linking the two.

Again, hope i helped . Now, probability questions tend to be tricky sometimes. Even i got confused writing this reply.

 

[muzirox]

Mutually exlusive events are like this. Suppose you have A and B. A is heads, B is tails. Now if you get a heads, the probability of tails is 0, and vice versa. So over here, A is a three on WHITE DICE. And B is a three on BLUE DICE. So if you get, for suppose, a three on the white dice, the probability of a three on blue dice is NOT 0, because the two dice are not related.
Consider this. A is a 3 on Blue Dice, and B is a 4 on Blue Dice. Now this is mutually exclusive, because only one can occur, that is, you can not get both. But in the question, you can get both, a 3 on Blue and a 3 on White, because there are two die. So it is NOT mutually exclusive. Hope that helped out.

Now, independent events are like this. If there are two events, they will be independent only if the probability of one of them occurring, does not affect the probability of the second one of them occurring. So, if you look over here, C and D can not be independent. This is because, the probability of a three on both dice, will reduce the probability of 4 on both dice. Over here, we are linking the two.

Again, hope i helped . Now, probability questions tend to be tricky sometimes. Even i got confused writing this reply.

Thanx a lotttttttttttttttttt

 

[muzirox]

y standardised death rate is better than crude death rate??


birth weight of babies------------either quantitative or qualitative

 

[Alchemist4lyf]

Standardised Death Rates take age standard populations into consideration as well. Crude Death rates take only age population. Thing is, Crude Death rate gives you, like, a summary type picture. Standardised death rates, give you like, the whole package. The whole picture. In it, the picture of better living conditions is given, for ALL ages. Like, if an old person wants to move to a new place, he'll look at that place's standardised death rate, not crude. Same is the case for the young person.

Weight can be expressed in numbers, so it is a quantitative variable. And, as it is MEASURED, it is a continuous quantitative variable.

 

[muzirox]

Thanx for the above & plzz brief me wat is cyclic & seasonal variation

 

[SalmanPakRocks]

Thanx for the above & plzz brief me wat is cyclic & seasonal variation

Cyclic: An approximate regular pattern, repeated is series of data, but often over different time periods.
Seasonal Variation: A pattern of fluctuation in a series of data which recurs at definete time intervals.

 

[hassankhan]

how do we find the height in a histogram plz help me.

plz solve 2010 Q6

 

[muzirox]

how do we find the height in a histogram plz help me.

plz solve 2010 Q6

which session??? which paper 12,22,13??????

 

[hassankhan]

2010 p11 Q 6