Math/Probability/Statistics? - canada education savings grant
Parents can help, with the exception of the college for their children's Registered Education Savings Plan (RESP). Under the Canada Education Savings Program of the Government meets 20% of parents contribute to an RESP (annual maximum). In addition, income earned in an RESP is not taxable (the amount of tax is deferred until the money is used for post-secondary).
In 2000, 7.2% of taxpayers with children under 19 years contributed to an RESP.
A primary school with 16 classes with 1 teacher and 20 students. While the parent-teacher interviews, ask any teacher, the parents of the students in his class if they contribute to an RESP for your child.
(a) What is the expected number of contributing parents (to a certain class).
(b) What is the probability that more than half of the parents to contribute data
Class?
(c) How large is the probability that 1 / 4 of the class to guarantee 3 on the relations of the parents?
Wednesday, December 30, 2009
Canada Education Savings Grant Math/Probability/Statistics?
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Let X be the number of parents who make an RESP account. X is the binomial distribution with n = 16 * 20 = 320 and the probability of success p = 0.072 has
In general, if X then with n trials and probability p for success, binomial
P [X = x] = n! / (X! (NX)!) * P ^ x * (1-p) ^ (NX)
For values of x = 0, 1, 2, ..., n
P [X = x] = 0 for all values of x.
This includes the analysis of the number of combinations of selected objects XnY be a great success and xn - x failures.
Or to be more accurate, the binomial is the sum of n independent and identically distributed Bernoulli trials.
the average of the binomial distribution is n * p
Variance of the binomial distribution is n * p * (1 - p)
contribute (a) The expected number of parents can be 320 * 0.072 = 23.04
(b) Let X be the number of parents contributing to the class. X ~ Binomial (n = 20, p = 0.072)
P (X> 10) = P (X = 11) + P (X = 12) + ... + P (X = 20)
= 2.452127e-08
(c) First, the probabilityspeed, the 3 or more to help parents in a class. Let X ~ binomial (20, 0,072), P (X> 3) = P (X = 4) + ... + P (X = 20) = 0.05140278
Be prepared Y is the number of classes to parents more than three taxpayers.
Y ~ Binomial (n = 16, p = 0.05140278)
P (Y = 4) = 0.00674528
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