Question

Let S16 = Σ Xiwhere {X1,X2, , X16} nd Geometric each with mean 2

• Find the mean of S₁₆.  
• Find the standard deviation of S₁₆. (Round it to one decimal place)
• Find P(S₁₆ > 40) using CLT, without correction factor.  (Round it to 4 decimal places)
• Find P(S₁₆ > 40) using CLT, with correction factor. (Round it to 4 decimal places
• FIND p0=exact = P(S₁₆ > 40). Note This is negative binomial with number of successes = n. Do not use Mathematica. It gives different answer because its definition of Negative Binomial is slightly different
• Is the given CLT an overestimate or underestimate or equal?

Answer 1

Let

Ai Geometric p

The expected value (Mean) of $\begin{align*} X_i \end{align*}$ is given by

E(X)=-= 2 → p = 0.5

Hence  
Xi ~ Geometric(p = 0.5)

The variance of $\begin{align*} X_i \end{align*}$ is

Var(K) = 1 Var(X) =i-p 1-0.5

Given that

16

The mean of is

S16

But we know that given 2 constants, a,b E(aX bY) aE(X) bE(Y) Extending this result to 16 variables we get - E(Xi) +E (X2) +... + E(X16) But each X, has a mean of 2 (They are iid)

The variance of is

S16

Var(S16) = Var (X1+ X2 + X16) But we know that given 2 constants, a,b and Independent X.Y Extending this result to 16 variables we get = Var(X) Var(X) + + Var(X16) X, has a variance of 2 (They are iid) But each = 32

The standard deviation of is

S16

SD(S16)- VVar(S16) V325.6569

Using CLT, we know that
S16
has a normal distribution with mean
μ=32
and standard deviation
σ 5.6569

The probability using CLT, without using the correction factor is

$\begin{align*} P(S_{16}>40)&=P\left(\frac{S_{16}-\mu}{\sigma}>\frac{40-\mu}{\sigma} \right )\quad\text{get the z score of 40}\\ &=P\left(Z>\frac{40-32}{5.6569} \right )\\ &=P(Z>1.41)\\ &=1-P(Z<1.41)\\ &=1-(0.5+0.4207)\quad\text{using the standard normal tables for z=1.41}\\ &=0.0793 \end{align*}$

The probability using CLT, with correction factor is

$\begin{align*} P(S_{16}>40)&=P(S_{16}>39.5)\quad\text{by adding the correction factor of -0.5}\\ &=P\left(\frac{S_{16}-\mu}{\sigma}>\frac{39.5-\mu}{\sigma} \right )\quad\text{get the z score of 40}\\ &=P\left(Z>\frac{39.5-32}{5.6569} \right )\\ &=P(Z>1.33)\\ &=1-P(Z<1.33)\\ &=1-(0.5+0.4082)\quad\text{using the standard normal tables for z=1.33}\\ &=0.0918 \end{align*}$

Each of Xs have 1 success, and hence can see that there are 16 successes in
S16
. That is
S16
is the number of trials required to get 16 successes, with probability of success on any given trial p=0.5.

We can say that has negative binomial distribution with parameters, number of successes =16 and the probability of success p=0.5

S16

The pmf of is

S16

$\begin{align*} P(S_{16}=s)&=\binom{s-1}{r-1}p^r(1-p)^{s-r}\\ &=\binom{s-1}{16-1}0.5^{16}(1-0.5)^{s-16}\\ &=\binom{s-1}{15}0.5^{s},\quad s=16,17,...\\ \end{align*}$

The exact probability is

P(S16 > 40) 1- P(S16 < 40) 39 s-16 39 0.5 15 s-16 0.5s 15!(s 1-15)! s-16 1- (0+0.00010.00050.0281) 1- 0.9002 - 0,0998

The CLT estimate with the correction is 0.0918 and without the correction is 0.0793.

Hence we can say that the CLT underestimates the exact probability.