*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.
To work through the given problem, we will go step by step:
(a) To find the mean of the distribution of the random variable Y, we multiply the probability of getting a tails (p = 0.5) by the number of tosses (n = 3600). Therefore, the mean is E(Y) = p*n = 0.5*3600 = 1800.
To find the standard deviation of the distribution, we need the variance, which is calculated as p(1-p) = 0.5(1-0.5) = 0.25. Then, the standard deviation (σ) is the square root of the variance, which is σ = sqrt(0.25) = 0.5.
(b) To find the chances that Y will fall between 1770 and 1830, we need to calculate the probabilities for each value and then sum them up. Since this is a binomial distribution, the probability of Y = x is given by the formula P(X=x) = (n choose x) * (p^x) * (1-p)^(n-x).
Using the given formula, we can calculate P(1771) and P(1831). Then, to find P(1770<=Y<=1830), we subtract P(1831) from P(1771).
(c) To find the chances that Y will fall between 1800 and 1830, we can follow the same steps as in part (b). Calculate P(1800) and P(1831), then subtract P(1831) from P(1800) to find P(1800<=Y<=1830).
(d) To find the chances that Y will fall between 1830 and 1860, repeat the steps from part (b). Calculate P(1831) and P(1861), then subtract P(1861) from P(1831) to find P(1830<=Y<=1860).
Note that when working with large numbers, direct evaluation of the probabilities using the formulas can be time-consuming or computationally challenging. In such cases, you might consider using statistical software or online calculators designed to handle these calculations efficiently.