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S3 chi-squared test Continuous Uniform Distribution question

hi(s3 page 88)
For Chi-squared test for continuous uniform distribution, do you still combine cells if expected frequency for a cell is <5?

On page 88 of the edexcel S3 book they dont seem to for the CUD question.

Is this a mistake or is there a different method for CUD?

Regards,
pbs
Original post by PolarBearSlayer
hi(s3 page 88)
For Chi-squared test for continuous uniform distribution, do you still combine cells if expected frequency for a cell is <5?

On page 88 of the edexcel S3 book they dont seem to for the CUD question.

Is this a mistake or is there a different method for CUD?

Regards,
pbs


Do you have a copy of the question for us to see?
Original post by Slowbro93
Do you have a copy of the question for us to see?



Hi, I think I've sorted the problem with that question, but I have a new S3 problem and would appreciate if you referred me to the relevant formula and explained what's going on:
7. A company produces climbing ropes. The lengths of the climbing ropes are normally
distributed. A random sample of 5 ropes is taken and the length, in metres, of each rope is
measured. The results are given below.
120.3 120.1 120.4 120.2 119.9

(a) Calculate unbiased estimates for the mean and the variance of the lengths of the

climbing ropes produced by the company.
(5)

The lengths of climbing rope are known to have a standard deviation of 0.2 m. The
company wants to make sure that there is a probability of at least 0.90 that the estimate
of the population mean, based on a random sample size of n, lies within 0.05 m of its true
value.
(b) Find the minimum sample size required.
(6)
_________


It's part b I can't do - would really appreciate some help. :smile:


PBS
Reply 3
Original post by PolarBearSlayer
Hi, I think I've sorted the problem with that question, but I have a new S3 problem and would appreciate if you referred me to the relevant formula and explained what's going on:
7. A company produces climbing ropes. The lengths of the climbing ropes are normally
distributed. A random sample of 5 ropes is taken and the length, in metres, of each rope is
measured. The results are given below.
120.3 120.1 120.4 120.2 119.9

(a) Calculate unbiased estimates for the mean and the variance of the lengths of the

climbing ropes produced by the company.
(5)

The lengths of climbing rope are known to have a standard deviation of 0.2 m. The
company wants to make sure that there is a probability of at least 0.90 that the estimate
of the population mean, based on a random sample size of n, lies within 0.05 m of its true
value.
(b) Find the minimum sample size required.
(6)
_________


It's part b I can't do - would really appreciate some help. :smile:


PBS


Well, since it's normal, life becomes a lot easier. Your sample mean follows a distribution of N(μ,σ2/n)N(\mu,\sigma^2/n). Standardise this as you would normally do and it should turn out fine! Best of luck
Original post by BananaPie
Well, since it's normal, life becomes a lot easier. Your sample mean follows a distribution of N(μ,σ2/n)N(\mu,\sigma^2/n). Standardise this as you would normally do and it should turn out fine! Best of luck



Thanks, a lot - seemed to work fine, I've got the hang of it!

Now, this question I could really do with a bit of help on

2. Philip and James are racing car drivers. Philip’s lap times, in seconds, are normally
distributed with mean 90 and variance 9. James’ lap times, in seconds, are normally
distributed with mean 91 and variance 12. The lap times of Philip and James are
independent. Before a race, they each take a qualifying lap.
(a) Find the probability that James’ time for the qualifying lap is less than Philip’s.
(4)
The race is made up of 60 laps. Assuming that they both start from the same starting line
and lap times are independent,
(b) find the probability that Philip beats James in the race by more than 2 minutes.

(a) Im fine on, but for b - when I do let y=(P1+..+P60)+(J1+...+J60), and find Y's normal distribution, I get a mean of -60 and standard deviation of root(1260). Hence, when finding P(Y>120), I end up with P(z>120--60/root(1260)) which gives me p(z>5.something), the wrong value.

In the ms,they do 120-60/root(1260) or 120--60/-root(1260).

Please explain why.

PBS
Original post by Slowbro93
Do you have a copy of the question for us to see?



Now, this question I could really do with a bit of help on

2. Philip and James are racing car drivers. Philip’s lap times, in seconds, are normally
distributed with mean 90 and variance 9. James’ lap times, in seconds, are normally
distributed with mean 91 and variance 12. The lap times of Philip and James are
independent. Before a race, they each take a qualifying lap.
(a) Find the probability that James’ time for the qualifying lap is less than Philip’s.
(4)
The race is made up of 60 laps. Assuming that they both start from the same starting line
and lap times are independent,
(b) find the probability that Philip beats James in the race by more than 2 minutes.

(a) Im fine on, but for b - when I do let y=(P1+..+P60)+(J1+...+J60), and find Y's normal distribution, I get a mean of -60 and standard deviation of root(1260). Hence, when finding P(Y>120), I end up with P(z>120--60/root(1260)) which gives me p(z>5.something), the wrong value.

In the ms,they do 120-60/root(1260) or 120--60/-root(1260).

Please explain why.

PBS
bump :smile:
Reply 7
Original post by PolarBearSlayer
Now, this question I could really do with a bit of help on

2. Philip and James are racing car drivers. Philip’s lap times, in seconds, are normally
distributed with mean 90 and variance 9. James’ lap times, in seconds, are normally
distributed with mean 91 and variance 12. The lap times of Philip and James are
independent. Before a race, they each take a qualifying lap.
(a) Find the probability that James’ time for the qualifying lap is less than Philip’s.
(4)
The race is made up of 60 laps. Assuming that they both start from the same starting line
and lap times are independent,
(b) find the probability that Philip beats James in the race by more than 2 minutes.

(a) Im fine on, but for b - when I do let y=(P1+..+P60)+(J1+...+J60), and find Y's normal distribution, I get a mean of -60 and standard deviation of root(1260). Hence, when finding P(Y>120), I end up with P(z>120--60/root(1260)) which gives me p(z>5.something), the wrong value.

In the ms,they do 120-60/root(1260) or 120--60/-root(1260).

Please explain why.

PBS


I get the same numbers as you.

I can see in the explanation of your working you let y=(P1+..P60) + (J1+..+J60) when it should be y=(P1+..P60) - (J1+..+J60)

Though I still get the same mean and s.d. as you. Where is this question from so I can look at it?
Reply 8


Care to explain?
Reply 10
Original post by PolarBearSlayer
hi(s3 page 88)
For Chi-squared test for continuous uniform distribution, do you still combine cells if expected frequency for a cell is <5?

On page 88 of the edexcel S3 book they dont seem to for the CUD question.

Is this a mistake or is there a different method for CUD?

Regards,
pbs


Can you please explain this? When I looked at the example of this one in the text book,I was so confused!
Thanks.

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