Further Maths statistics question
A lottery has tickets numbered 1 to n inclusive, where n is a positive integer. The random variable
X denotes the number on a ticket drawn at random.
(a) Determine P( X ≤ 0.25n)
in each of the following cases.
(i) n is a multiple of 4. [1]
(ii) n is of the form 4k + 1 where k is a positive integer. Give your answer as a single fraction
in terms of n.
answer from mark scheme: (for (ii))
P(X≤0.25n) = k/(4k + 1)
= (0.25(n – 1))/n
= (n–1)/4n
// i dont really understand the second question in terms of how they got the answer (like first and second part) and i couldnt really find any similar questions/explanations on this in the ocr b textbook or anywhere else.
I believe its a discrete uniform distribution question but anyway I'd really appreciate it if someone was able to answer soon (i have my stats mock on wednesday ðŸ˜)
X denotes the number on a ticket drawn at random.
(a) Determine P( X ≤ 0.25n)
in each of the following cases.
(i) n is a multiple of 4. [1]
(ii) n is of the form 4k + 1 where k is a positive integer. Give your answer as a single fraction
in terms of n.
answer from mark scheme: (for (ii))
P(X≤0.25n) = k/(4k + 1)
= (0.25(n – 1))/n
= (n–1)/4n
// i dont really understand the second question in terms of how they got the answer (like first and second part) and i couldnt really find any similar questions/explanations on this in the ocr b textbook or anywhere else.
I believe its a discrete uniform distribution question but anyway I'd really appreciate it if someone was able to answer soon (i have my stats mock on wednesday ðŸ˜)
Original post by chericheri
A lottery has tickets numbered 1 to n inclusive, where n is a positive integer. The random variable
X denotes the number on a ticket drawn at random.
(a) Determine P( X ≤ 0.25n)
in each of the following cases.
(i) n is a multiple of 4. [1]
(ii) n is of the form 4k + 1 where k is a positive integer. Give your answer as a single fraction
in terms of n.
answer from mark scheme: (for (ii))
P(X≤0.25n) = k/(4k + 1)
= (0.25(n – 1))/n
= (n–1)/4n
// i dont really understand the second question in terms of how they got the answer (like first and second part) and i couldnt really find any similar questions/explanations on this in the ocr b textbook or anywhere else.
I believe its a discrete uniform distribution question but anyway I'd really appreciate it if someone was able to answer soon (i have my stats mock on wednesday ðŸ˜)
X denotes the number on a ticket drawn at random.
(a) Determine P( X ≤ 0.25n)
in each of the following cases.
(i) n is a multiple of 4. [1]
(ii) n is of the form 4k + 1 where k is a positive integer. Give your answer as a single fraction
in terms of n.
answer from mark scheme: (for (ii))
P(X≤0.25n) = k/(4k + 1)
= (0.25(n – 1))/n
= (n–1)/4n
// i dont really understand the second question in terms of how they got the answer (like first and second part) and i couldnt really find any similar questions/explanations on this in the ocr b textbook or anywhere else.
I believe its a discrete uniform distribution question but anyway I'd really appreciate it if someone was able to answer soon (i have my stats mock on wednesday ðŸ˜)
Since you've done the first part, then you know the general method.
For part ii, 0.25n is k+ (1/4).
The probabilty that X <=0.25n is the probability that X <=k +(1/4).
But as you note, we are dealing with a discrete distribution, the R.V. only takes integer values; it can't take values between k and k+(1/4)
So, P(X<= k+(1/4)) = P(X<=k) and the rest follows.
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