# Math

posted by
**SherylA** on
.

*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.