Simple STATISTICS 1 PROBLEM...really stuck!

Question on Probability.

A six-sided die is biased such that there is an equal chance of scoring each of the numbers from 1 to 5 but a score of 6 is three times more likely than each of the other numbers.
(a) Write down the probability distribution for the random variable, X, the score on a single throw of the die.

Reply 3

Eloades11

17

Original post by pHneutral

Hi guys,

I am really stuck here, please help me out.
I am doing S1 edexcel and my textbook states the below:

In the normal distribution chapter, (page 179)

If P ( Z< a ) is greater than 0.5, then a will be > 0
If P ( Z< a ) is less than 0.5, then a will be less than 0

If P ( Z > a ) is LESS than 0.5, then a will be > 0
If P ( Z > a ) is more than 0.5, then a will be < 0

1. I don't understnad how the normal distribution diagram for each of the above senarios looks like.
2. The book expects that in future questions based on the above one should be able to draw the diagrams, but I don't get how you would know exactly where the 'a' is for instance, like is it positive or negative side of mean or anything.
3. EXAMPLE QUESTION: P (Z > a) = 0.6879 . How does the sketch for this one look like?
4. Are there any tricks or techniques that I am missing, which could be helpful to me in understanding the above.

I am homestudying, cannot ask a teacher or anything :frown:

Really really appreciate any little help.
Thanks

Basically the normal distribution has a mean of 0 and a variance of 1, this gives
Z~N(0,1)

lets do your example, I'm assuming you have to find 'a' in p(Z>a).
Firstly we can't deal with anything when it is Z>... because the values in the table are cumulative, so we can change this to p(Z>a) = 1 - p(Z< or = a) = 0.6879
This gives p(Z<a) = 0.3121
Attached below is my quick/rough sketch on paint.

You know that on each side of the mean has a probability of 0.5, as it is symmetrical. So because it is cumulative, as the positive x value tends to infinite, the cumulative value tends to 1.

I sketched the graph of the region for p(Z<a) = 0.3121
I'll let you try to sketch p(Z>a), if you get stuck check the spoiler

Spoiler

If you don't understand then just ask :smile:

EDIT: Beaten to it :tongue:

NormalDistr..jpg18.3KB

Reply 4

OP

Original post by al_habib

Question on Probability.

A six-sided die is biased such that there is an equal chance of scoring each of the numbers from 1 to 5 but a score of 6 is three times more likely than each of the other numbers.
(a) Write down the probability distribution for the random variable, X, the score on a single throw of the die.

Below is my working out:

X [ 1 2 3 4 5 6 ]

p(X =x ) [ x/6 x/6 x/6 x/6 x/6 3times x/6 ]

USE an unknown variable , x

All probabilities must add up to 1.

Therefore, x/6 + x/6 + x/6 + x/6 + 0.5x = 1

8x/6 =1
x= 0.75

Now fill in the table with x replaced.

Do you have the answers??

(edited 12 years ago)

Reply 5

another que.

i did part a & b

The individual letters of the word STATISTICAL are written on 11 cards which are then shuffled.
One card is selected at random. Find the probability that it is
(a) a vowel, (1 mark)
(b) a T, given that it is a consonant. (2 marks)
The 11 cards are then shuffled again and the top three are turned over. Find the probability that
(c) all three of the cards have a T on them, (3 marks) (d) at least two of the cards show a vowel. (6 marks)

Reply 6

OP

Original post by al_habib

....

Original post by Eloades11

.....

If you don't understand then just ask :smile:

EDIT: Beaten to it :tongue:

thanks for the help alot.

If I may ask the below question:

The random variable X ~ N(20, 12).

Find the value of a and the value of b such that

a P(X < a) = 0.40,

b P(X > b) = 0.6915.

c Write down P(b < X < a).

For part a) I dont understand why (a-20)/ root 12 = -0.2533, why is it negative and not positive?
Part b) I am stuck.

I am posting the anwers which i dont understnad:
Solution:
(a) p(X<a) =0.40 UseP=0.4000
a?20 \?12 =?0.2533
a =19.122…?a=19.1(3sf)

(b)
P(X>b) =0.6915
b?20 \?12 =?0.5
?b =18.267…?b=18.3(3sf)

(c)p(b<X<a)
=0.40?[1?0.6915]
=0.0915

thanks

Reply 7

Original post by pHneutral

Below is my working out:

X [ 1 2 3 4 5 6 ]

p(X =x ) [ x/6 x/6 x/6 x/6 x/6 3times x/6 ]

USE an unknown variable , x

All probabilities must add up to 1.

Therefore, x/6 + x/6 + x/6 + x/6 + 0.5x = 1

8x/6 =1
x= 0.75

Now fill in the table with x replaced.

Do you have the answers??

there you go

x 1 2 3 4 5 6

P(X=x) 1/8 1/8 1/8 1/8 1/8 3/8

i wonder y they are using /8 instead of /6

Reply 8

OP

Original post by al_habib

there you go

x 1 2 3 4 5 6

P(X=x) 1/8 1/8 1/8 1/8 1/8 3/8

i wonder y they are using /8 instead of /6

They are not using 1/8,
that ends up being the answer. The are 6 different outcomes but all dont have same chance of being selected hence out of 6 is not used.

If you look at how I did it, you will also end up with the correct answers, using algebra makes it much easier.

Reply 9

Arsey

10

if you have at this thread I have uploaded some revision notes including normal distribution.

http://www.thestudentroom.co.uk/showthread.php?p=31185025

Reply 10

Eloades11

17

Original post by pHneutral

thanks for the help alot.

If I may ask the below question:

The random variable X ~ N(20, 12).

Find the value of a and the value of b such that

a P(X < a) = 0.40,

b P(X > b) = 0.6915.

c Write down P(b < X < a).

For part a) I dont understand why (a-20)/ root 12 = -0.2533, why is it negative and not positive?
Part b) I am stuck.

I am posting the anwers which i dont understnad:
Solution:
(a) p(X<a) =0.40 UseP=0.4000
a?20 \?12 =?0.2533
a =19.122…?a=19.1(3sf)

(b)
P(X>b) =0.6915
b?20 \?12 =?0.5
?b =18.267…?b=18.3(3sf)

(c)p(b<X<a)
=0.40?[1?0.6915]
=0.0915

thanks

Sorry I didn't understand if you solved this in the last posts.

You have X~N(20,12)
You want to find 'a' such that p(X<a)= 0.4000

First you standardise the X variable using the equation Z=(X-mean)/SD

p(X<a) = p[Z<(a-20)/root(12)] =0.4000
You know that p = 0.4, so you can find z in the tables. Then once you found this value(let this value = x) Now you can find 'a' by doing (a-20)/root12 = x

Similar sort of thing with your next question, although remember what I said about Z being greater than, you have to convert it to 1-p(Z<....

Reply 11

Original post by pHneutral

thanks for the help alot.

If I may ask the below question:

The random variable X ~ N(20, 12).

Find the value of a and the value of b such that

a P(X < a) = 0.40,

b P(X > b) = 0.6915.

c Write down P(b < X < a).

For part a) I dont understand why (a-20)/ root 12 = -0.2533, why is it negative and not positive?
Part b) I am stuck.

I am posting the anwers which i dont understnad:
Solution:
(a) p(X<a) =0.40 UseP=0.4000
a?20 \?12 =?0.2533
a =19.122…?a=19.1(3sf)

(b)
P(X>b) =0.6915
b?20 \?12 =?0.5
?b =18.267…?b=18.3(3sf)

(c)p(b<X<a)
=0.40?[1?0.6915]
=0.0915

thanks

(a) get sorted with 0.40 1st, draw your diagram 0.40 is less than 0.5 it will be -0.40 you change into +0.40 and this P(X<a) changes to P(X>a) :. take 1-P(X<x) . always we use P(X<x), which is going to be 1-0.40=0.60 read the table which should give you 0.25 since you change < to > ans -0.25

we use root 12 to get standard deviation we cant use variance 12 squared.

part b is just the same.

Reply 12

Eloades11

17

Original post by al_habib

(a) get sorted with 0.40 1st, draw your diagram 0.40 is less than 0.5 it will be -0.40 you change into +0.40 and this P(X<a) changes to P(X>a) :. take 1-P(X<x) . always we use P(X<x), which is going to be 1-0.40=0.60 read the table which should give you 0.25 since you change < to > ans -0.25

we use root 12 to get standard deviation we cant use variance 12 squared.

part b is just the same.

Sorry, how does that help you find a?

Reply 13

Original post by Eloades11

Sorry, how does that help you find a?

applyy the formula..

(edited 12 years ago)

Reply 14

OP

Question:
The random variable X ~ N(mean, sigma^2)
The lower quartile of X is 25 and the upper quartile of X is 45.
Find the value of mean , and the value of sigma

I dont understand how one would do this question systematically?
thanks!

(edited 12 years ago)

Reply 15

OP

For the above, why is the proability 0.75 in the mark schme.
Why not 0.5?

Reply 16

lower quartile, 1/4=0.25 upper quartile,3/4=0.75

P(X<25)=0.25
P(X<45)=0.75

then use the formula Z = X-mean/SD

:. 25 - mean/ SD =0.25
& 45 - mean/ SD =0.75

then solve simultaneously

(edited 12 years ago)

Reply 17

PROBABILITY PROBLEM

In a study of 120 pet-owners it was found that 57 owned at least one dog and of these 16 also owned at least one cat. There were 35 people in the group who didn?’t own any cats or dogs.

As an incentive to take part in the study, one participant is chosen at random to win a year?’s free supply of pet food.

Find the probability that the winner of this prize

(a) owns a dog but does not own a cat, (2 marks)

(b) owns a cat, (4 marks)

(c) does not own a cat given that they do not own a dog. (4 marks)

Reply 18