Question

I need the calculation for these probability and statistics question...

Question 1. Faid decided to perform independent Bernoulli(p) trials and stop immediately after the 115th trial. Suppose that he observed exactly 20 successes in those 115 trials. Estimate p.

Answer: 0.1739   Hint: The sample mean is an unbiased estimator for p.

Question 2 : Faid decided to perform independent Bernoulli(p) trials and stop immediately after his 19th success. Suppose that his 19th success was on the 170th trial. Estimate p.

Answer: 0.1065   Hint: The fraction of successes prior to the last trial is an unbiased estimator for p.​​​​​​​

Answer 1

Question 1:

let , the population mean is P which is the probability of success.

now faid decided to perform independent bernoulli(p) where p is the probability of success.

we draw a random sample from the population which is x₁ ,x₂ , ....... , x_{n .}

here, xi, (i=1 to n) are random variables which denote the result of i th trial.

now, xi~ber(p) , i=1 to n

now . faid stop the 115th trail and observed exactly 20 successes in those 115 trail.

then here n=115. and no of success is s=20.

so. the probability of success is p=s/n=(20/115)= 0.1739.

now the sample mean is p=0.1739 [as, xi~ber()p; i=1 to n]

now, sample mean is an unbiased estimator of P which is population mean ; i.e: E(p)=P

[proof:

let $\overline{x}$ , the mean of x_{1, .....,} x_n randomly selected sample from the population with mean m.

so. E($\overline{x}$)=E((x₁ + x₂ +......+x₂)/n)= 1/n {E(x₁)+E(x₂)+.....+E(x_n)} = nm/n=m]

so here,estimated , P =E(p)=0.1739... (Ans)

Question 2:

let x_1,x_{2, ........ ,} x_n be the random sample selected from the population with mean P [as xi~ber(p)]

so E(x_i)=P

here xi denote the result of ith trial.

let , xi = 1, if success occur in ith trial

0, if failure occur in ith trial.  

then the sample mean is p. now the fraction of success prior to the last trial is

I – u/ = [d

is the no of success in n-1 trial.

n-1 where, Στί, i=1

now,

E(p1)= 1/(n-1)[E(x₁+x₂+ .... +x_n-1)]= 1/(n-1)[E(x₁)+E(x₂)+ ..... +E(x_n-1)] = [(n-1)P]/(n-1)=P

here no of success in 169 th trial is 18. because faid stop the trials immediately after his 19th success which was on 170th trial.

so the fraction of success prior to the last trail is p1= 18/169 =0.1065.

now , p1 is an unbiased estimatior of P. [as, E(p1)=P]

so , the esimated value of P is 0.1065. ....(Ans)