S2 Mathematics : Masters of statistics 2 please help us

[kirashinagami]

So, paper 7 will be on 29th so we must unite and tackle the problems together. Luckily I have been able to do 3 or 4 questions out of 6 or 7 questions in the past papers. So, I'm hoping that I may clear your doubts and I'll do my best to clear yours so, let's start! I'm gonna ask first.

 

[kirashinagami]

The weights of bags of fuel have mean 3.2 kg and standard deviation 0.04 kg. The total weight of a random sample of three bags is denoted by T kg. Find the mean and standard deviation of T .

Why the mean is 3.2 x 3 ? Why should it be multiplied by 3 ?
Why the sd is square root of (3 x (0.04)^2) ?

 

[kirashinagami]

The number of goals scored per match by Everly Rovers is represented by the random variable X which has mean 1.8.
(i) State two conditions for X to be modelled by a Poisson distribution.
Assume now that X ∼ Po(1.8).
(ii) Find P(2 < X < 6).
(iii) The manager promises the team a bonus if they score at least 1 goal in each of the next 10 matches. Find the probability that they win the bonus.

I only didn't understand the part 3 , what is the concept behind it? The answer is ( 1 - e^-1.8)^10

 

[kirashinagami]

The number of adult customers arriving in a shop during a 5-minute period is modelled by a random variable with distribution Po(6). The number of child customers arriving in the same shop during a 10-minute period is modelled by an independent random variable with distribution Po(4.5).
(i) Find the probability that during a randomly chosen 2-minute period, the total number of adult and child customers who arrive in the shop is less than 3. [3]
(ii) During a sale, the manager claims that more adult customers are arriving than usual. In a randomly selected 30-minute period during the sale, 49 adult customers arrive. Test the manager’s claim at the 2.5% significance level. [6]

The question is easy, what I didn't understand is in part 2 why we take 48.5 for continuity correction? Why not 49.5?

 

[kirashinagami]

Jeevan thinks that a six-sided die is biased in favour of six. In order to test this, Jeevan throws the die 10 times. If the die shows a six on at least 4 throws out of 10, she will conclude that she is correct.
(i) State appropriate null and alternative hypotheses.
(ii) Calculate the probability of a Type I error.
(iii) Explain what is meant by a Type II error in this situation.
(iv) If the die is actually biased so that the probability of throwing a six is 1 , calculate the probability 2 of a Type II error.

I'm so weak this. Can anyone explain me?

 

[kirashinagami]

Now ask me your doubts , I'll help to solve it!

 

[kirashinagami]

Hello

 

[kirashinagami]

Help

 

[soul1]

Hey

 

[redapple20]

http://www.xtremepapers.com/communi...-2-post-your-doubts-and-solve-mine-lol.14039/

 

[kirashinagami]

P7 is near. Let's start discussing!

 

[gudboy01]

The weights of bags of fuel have mean 3.2 kg and standard deviation 0.04 kg. The total weight of a random sample of three bags is denoted by T kg. Find the mean and standard deviation of T .

Why the mean is 3.2 x 3 ? Why should it be multiplied by 3 ?
Why the sd is square root of (3 x (0.04)^2) ?

Becoz the mean n sd given in que is for one bag but it is asked to find for 3 bags...
E(3B)=3xE(B)=3x302
For var(3B)=3xVar(B)=3*0.04^2 [all the bags r identical sooo dont square the 3]

 

[gudboy01]

The number of goals scored per match by Everly Rovers is represented by the random variable X which has mean 1.8.
(i) State two conditions for X to be modelled by a Poisson distribution.
Assume now that X ∼ Po(1.8).
(ii) Find P(2 < X < 6).
(iii) The manager promises the team a bonus if they score at least 1 goal in each of the next 10 matches. Find the probability that they win the bonus.

I only didn't understand the part 3 , what is the concept behind it? The answer is ( 1 - e^-1.8)^10

I think its already done to u

 

[gudboy01]

The number of adult customers arriving in a shop during a 5-minute period is modelled by a random variable with distribution Po(6). The number of child customers arriving in the same shop during a 10-minute period is modelled by an independent random variable with distribution Po(4.5).
(i) Find the probability that during a randomly chosen 2-minute period, the total number of adult and child customers who arrive in the shop is less than 3. [3]
(ii) During a sale, the manager claims that more adult customers are arriving than usual. In a randomly selected 30-minute period during the sale, 49 adult customers arrive. Test the manager’s claim at the 2.5% significance level. [6]

The question is easy, what I didn't understand is in part 2 why we take 48.5 for continuity correction? Why not 49.5?

becoz here alternative hypothesis > 36 so we hab to find P(X greater or equal to 49).
here lamda is 36 which is greater than 15 so valid normal approximation
then P(Y>48.5) [as there is greater than or equal to 49 we cant take 49.5 becoz 49 is not greater than 49.5 but 49 is greater than 48.5]
Hope u understand it...

 

[gudboy01]

Jeevan thinks that a six-sided die is biased in favour of six. In order to test this, Jeevan throws the die 10 times. If the die shows a six on at least 4 throws out of 10, she will conclude that she is correct.
(i) State appropriate null and alternative hypotheses.
(ii) Calculate the probability of a Type I error.
(iii) Explain what is meant by a Type II error in this situation.
(iv) If the die is actually biased so that the probability of throwing a six is 1 , calculate the probability 2 of a Type II error.

I'm so weak this. Can anyone explain me?

i) null P(6)= 1/6
alternative P(6) > 1/6 [becoz die is bias in favor of 6, soo prob of getting 6 is greater than 1/6]
ii) here n=10 becoz its thrown 10 times
type I error means reject Ho or accept H1
soo here u hab to find P(X greater than equal to 4)
find it using binomial distribution with p=1/6
iii) say that prob of getting 6 is 1/6 when its actually more than 1/6 [its becoz type 2 error means accepting Ho when Ho is false]
iv) here its given p=1 and again n=10
type II error means accept Ho or reject H1
soo here u hab to find P(X less than 4)
find it using binomial again...
HOPE IT WILL HELP U AND ALL THE BEST FOR TOMORROW'S EXAM....

 

[ALstudentd]

Hi, could anyone kindly clear my doubts in the following questions:

M/J 2010 P 71:
In Q2 part ii) why do we equate half width with z(root(p*q/n))
In Q 4 part ii) why should we take it as E(5/8X) and Var(5/8X) rather than finding in km first then converting to miles
In Q 5 part iii) why is the integral equated to 0.2 and not 0.8

M/J 2011 P 71:
Q2 part b) why is (1/16) multiplied by 2 ? I am assuming it is a one-tailed test, since the question states that the interval contains values lower than population mean.
Q3 part ii) I didn't understand why we compare the given test statistic to the z value for 5% significance level

 

[gudboy01]

Hi, could anyone kindly clear my doubts in the following questions:

M/J 2010 P 71:
In Q2 part ii) why do we equate half width with z(root(p*q/n))
In Q 4 part ii) why should we take it as E(5/8X) and Var(5/8X) rather than finding in km first then converting to miles
In Q 5 part iii) why is the integral equated to 0.2 and not 0.8

M/J 2011 P 71:
Q2 part b) why is (1/16) multiplied by 2 ? I am assuming it is a one-tailed test, since the question states that the interval contains values lower than population mean.
Q3 part ii) I didn't understand why we compare the given test statistic to the z value for 5% significance level

mj10 p71
2)ii) its formula that is width = 2*z*root(p*q/n) so width is divided by 2.
4)ii) it can b done as u said..1st find the combined mean and s.d of (mr+mrs) in km which u will get (512+89) = 601 and (62^2+7.4^2) =3898.76..now u can convert it into miles mean = 5/8 of 601= 375.63 and s.d = (5/8)^2*3898.76=1522.95
5)iii) i sud say dorry for dis.no idea for it

 

[gudboy01]

Hi, could anyone kindly clear my doubts in the following questions:

M/J 2010 P 71:
In Q2 part ii) why do we equate half width with z(root(p*q/n))
In Q 4 part ii) why should we take it as E(5/8X) and Var(5/8X) rather than finding in km first then converting to miles
In Q 5 part iii) why is the integral equated to 0.2 and not 0.8

M/J 2011 P 71:
Q2 part b) why is (1/16) multiplied by 2 ? I am assuming it is a one-tailed test, since the question states that the interval contains values lower than population mean.
Q3 part ii) I didn't understand why we compare the given test statistic to the z value for 5% significance level

for mj 11 p71,
2)b)again sorry for dis..
3)ii) we hab to test with z value of 5%significance level to know the rejection of acceptance of Hp..
At 5% signi. level,
z=1.645
Given test stat, z=1.786..
Hence, reject Ho coz 1.786> 1.645..

 

[ALstudentd]

for mj 11 p71,
2)b)again sorry for dis..
3)ii) we hab to test with z value of 5%significance level to know the rejection of acceptance of Hp..
At 5% signi. level,
z=1.645
Given test stat, z=1.786..
Hence, reject Ho coz 1.786> 1.645..

gudboy01, thank you so for your help in both questions

 

[ALstudentd]

gudboy01, thank you so for your help in both questions

so much*

I also have one more question, if you could please explain,
M/J 2011 p73
Q2 part i) Firstly, isn't the approximation from binomial to normal taken as (n*p, n*p*(1-p))? The examiner report states that it is (p, [p*(1-p)]/n). Please clarify this.
and secondly, from the question, I concluded that it is one-tailed, as the question mentions bars of soap being undersized, however, the z value used in the MS was for a two-tailed test ( 1.645 was used, not 1.282 for the 90% C.I.) :/

and in questions like Q3 part i), i don't know which proportions to use in my binomial calculation ( in this case 0.15 and 0.85 were the original proportions, but i used 2/30 and 28/30 )