February 19, 2017

Homework Help: Math

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

*Sigh* Can someone explain to me how I would work through 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.

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