Tuesday
August 4, 2015

Homework Help: Math

Posted by SherylA on Thursday, October 5, 2006 at 7:01pm.

*Sigh* Can someone explain to me how I would work thru this problem?

An honest coin tossed n=3600 times. Let the random variable Y denote the number of tails tossed.
(a) find the mean and the standard deviation of the distribution of the random variable Y
(b) what are the chances that Y will fall comewhere between 1770 and 1830?
(c) what are the chances that Y will fall somewhere between 1800 and 1830?
(d) what are the chances that Y will fall somewhere between 1830 and 1860?


By honest coin we mean one where P(H)=P(T)=.5 with H and T standing for heads and tails.
This is a Bernoulli trial with mean p=.5 variance = p(1-p)=.5^2=.25 and standard deviation = sqrt(variance)=.5
To answer (a) the expected number of tails is p*n where n is the number of tosses, so .5*3600 is E(#Tails). The s.d should still be .5 since that's the prob. of tails and heads.
For (b) P(1770<=X<=1830) so use P(X=x) where x is a number of trials. The prob. of P(x) = [n choose x]*(p^x)(1-p)^(n-x). The P(X<=1830)=1-P(1831) and P(1770<=X)=1-P(1771), so P(1770<=X<=1830) is P(1771) - P(1831).
Considering the size of the numbers you won't be able to evaluate them directly, unless you take the time to do some arithmetic.
Parts (c) and (d) are done similarly.

Answer this Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Members