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# S2 edexcel question- really need help watch

1. a large dental practice wishes to investigate level of satisfaction of paitient

a)suggest suitable sampling frame
b)identify sampling units
c)state one disadv and one adv of using a sample survey rather than a census
d) suggest a problem that might arise with the sampling frame when selecting patients

also
why might the population and the sampling frame not be the same?
give an example of when you might use:
a sampe
a census

thanx
2. (Original post by perkyDani)
a large dental practice wishes to investigate level of satisfaction of paitient

a)suggest suitable sampling frame
b)identify sampling units
c)state one disadv and one adv of using a sample survey rather than a census
d) suggest a problem that might arise with the sampling frame when selecting patients

also
why might the population and the sampling frame not be the same?
give an example of when you might use:
a sampe
a census

thanx
here is what i think
a)a list of registered paitents
b) the paitents
c)dis - not as accurate, eg bias
d) a lot of paitents may not respond

they can not be the same eg when the sampling frame changes such as paitents move house or practise maybe?
3. (Original post by perkyDani)
a large dental practice wishes to investigate level of satisfaction of paitient

a)suggest suitable sampling frame
b)identify sampling units
c)state one disadv and one adv of using a sample survey rather than a census
d) suggest a problem that might arise with the sampling frame when selecting patients
(a) The list of patients registered at the practice.
(b) The patients.
(d) Some patients on the list could have moved away or died. Some might not be willing to answer questions.
why might the population and the sampling frame not be the same?
Say your population is "Customers of WH Smith". There is no list of such people, so you have to choose a smaller sampling frame - for example, the customers of a particular branch on a particular day.
4. 1. The random variables R, S and T are distributed as follows R ∼ B(15, 0.3), S ∼ Po(7.5), T ∼ N(8, 22).
Find
(a) P(R = 5),
(b) P(S = 5),
(c) P(T = 5).

can someone help me with part (c) plz i really need help.
5. (Original post by nageen)
1. The random variables R, S and T are distributed as follows R ~ B(15, 0.3), S ~ Po(7.5), T ~ N(8, 22).
Find
(a) P(R = 5),
(b) P(S = 5),
(c) P(T = 5).

can someone help me with part (c) plz i really need help.
Short answer (enough for the exam):
0, because T is a continuous RV.

If X is a continuous RV then, for any a and b with a <= b,

P(a <= X <= b) = f(x) dx

where f is the pdf of X. So

P(X = a)
= P(a <= X <= a)
= f(x) dx
= 0

because the area represented by the integral is 0.
6. standardis 5-8 over root 22

then 1 minus the fi answer from the table

i get 0.2643
7. (Original post by Jonny W)
Short answer (enough for the exam):
0, because T is a continuous RV.

If X is a continuous RV then, for any a and b with a <= b,

P(a <= X <= b) = f(x) dx

where f is the pdf of X. So

P(X = a)
= P(a <= X <= a)
= f(x) dx
= 0

because the area represented by the integral is 0.
um its normally distributed matey
8. (Original post by Finn)
um its normally distributed matey
A normal distribution is an example of a continuous distribution.
9. er no your answer is wrong so is mine though i didnt realise it was just 5, though less than or equal to, if its just five its the probability less than or equal to 5 minus probability less than or equal to 4.
10. (Original post by Finn)
er no your answer is wrong so is mine though i didnt realise it was just 5, though less than or equal to, if its just five its the probability less than or equal to 5 minus probability less than or equal to 4.
What is the probability that a randomly selected person is EXACTLY six feet tall?
11. i get 0.2643 minus 0.1977 = 0.0666 i think thats the right answer
12. no P T=5 is greater than 4.5 and less than5.5 using continuity so there is a probability as ive worked out

0.0666
13. although i do see what your saying. id assume they are asking you to use continuity because they wouldnt aske for proof answers like that? or am i wrong...:-s
14. (Original post by Finn)
i get 0.2643 minus 0.1977 = 0.0666 i think thats the right answer
p(t=5)=0
15. lol how dare you ppl not listen to Jonny. for the person that said P(X<=5)-P(X<=4), you seem to be forgetting that this also includes ppl who are 4.5 (feet?) tall. The answer must be zero because the probability is infinitessively(my very own word) small
16. ok sorry just ive been working on assuming that any normal distribution work in s2 is done using continuity correction. i apologise.
17. (Original post by Finn)
ok sorry just ive been working on assuming that any normal distribution work in s2 is done using continuity correction. i apologise.
Don't worry. It's fun for me to have someone to argue with.

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