# Simple STATISTICS 1 PROBLEM...really stuck!Watch

#1
Hi guys,

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

Really really appreciate any little help.
Thanks
0
7 years ago
#2
(Original post by pHneutral)
Hi guys,

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

Really really appreciate any little help.
Thanks
I am in that situation as well but am planning to c one of my previous teacher for questions on probabilty, i'll post the question here 1st and c if i can get any ideas.

regarding your eg. Question: P(Z>a)=0.6879 since 0.6879 is greater than 0.5 then it is less than 0 it means it is -a.
method to follow:

1st- draw the normal distribution diagram allocate your a which is going to be less than 0 in this case -0.6879 on the left side of the mean.

2nd- switch -0.6879(-a) into the positive side it will be 0.6879(+a) then read this figure in normal distribution table P(Z<z) which is 0.49

3rd- since you changed the sign from > to < or from -ve a to +ve a then your ans becomes -0.49

put your head down practice as many questions in the text book as you can, try to be patience.

hope that helps, if you have questions or you dont understand any where feel free to ask.
1
7 years ago
#3
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.
0
7 years ago
#4
(Original post by pHneutral)
Hi guys,

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

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:
Show
basically the same as mine, but the shaded region is on the other side of a.

If you don't understand then just ask

EDIT: Beaten to it
0
#5
(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.

0
7 years ago
#6
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)
0
#7
(Original post by al_habib)
....
.....[/spoiler]

If you don't understand then just ask

EDIT: Beaten to it
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
0
7 years ago
#8
(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.

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
0
#9
(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.
0
7 years ago
#10
if you have at this thread I have uploaded some revision notes including normal distribution.

http://www.thestudentroom.co.uk/show...php?p=31185025
7 years ago
#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
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<....
0
7 years ago
#12
(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.
0
7 years ago
#13
(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.
0
7 years ago
#14
applyy the formula..
0
#15
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!
0
#16
For the above, why is the proability 0.75 in the mark schme.
Why not 0.5?
0
7 years ago
#17
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
0
7 years ago
#18
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)
0
#19
Really stuck!! Got exam tomarrow...

7. A packing plant fills bags with cement. The weight X kg of a bag of cement can be
modelled by a normal distribution with mean 50 kg and standard deviation 2 kg.
(a) Find P(X >53).
(b) Find the weight that is exceeded by 99% of the bags.
Three bags are selected at random.
(c) Find the probability that two weigh more than 53 kg and one weighs less than 53 kg.

I dont understand part C. I get part A AND B.
Mark scheme says :

P(2 weigh more than 53kg and 1 less) = 3×0.0668^2 times (1? 0.0668) = 0.012492487..

Why is 0.0668 times by 3?
And why is it being squared?

0
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