A coin was flipped 5 times end the number of heads recorded

A coin was flipped 5 times end the number of heads recorded after each toss was recorded. What is \"n\" the number of trials? What is \"p\" (probability of success) What is \"q\" (probability of failure) What is the probability that all five tosses produced a head? What is the probability that at least one toss was a head? What is the expected value (mean) of this distribution? What is the standard deviation of this distribution?

Solution

4. a coin was flipped 5 times and the number of heads recorded after each toss was recorded.

a) so here n=number of trials=number of times the coin was flipped=5 [answer]

b) p=probability of success=probability of obtaining head=1/2 [as there are only two outcomes \"head\" or \"tail\"] [answer]

c) q=probability of failure=1-p=1-1/2=1/2   [answer]

let X denote the total number of heads obtained.

so here X follows a Binomial distribution with parameters n=5 and p=0.5

so X~Bin(5,0.5)

so pmf of X is P[X=x]=⁵C_x0.5^x(1-0.5)^5-x=⁵C_x0.5⁵   x=0,1,2,3,4,5

d) so P[all five tosses produced a head]=P[X=5]=⁵C₅0.5⁵=0.03125 [answer]

e) P[ at least one toss was a head]=P[X>=1]=1-P[X=0]=1-⁵C₀0.5⁵=1-0.03125=0.96875 [answer]

f) X follows a Binomial distribution with parameters 5 and 0.5

now for a Binomial (n,p) distribution the expected value is np

here n=5 p=0.5

so the expected value(mean) of the distribution is

E[X]=5*0.5=2.5   [answer]

g) for a binomial (n,p) distribution the variance is np(1-p)

here n=5 p=0.5

so the variance is V[X]=5*0.5*0.5=1.25

hence the standard deviation of the distribution is sqrt(V[X])=sqrt(1.25)=1.118 [answer]