DATA 605 Week09 Discussion

CENTRAL LIMIT THEOREM (Chapter 9 - Section 9.1) - Question 10

Solution:

We know that total number of digits is 10. So, the probability of getting digit 3 is 1/10.
\(p = \frac{1}{10} = 0.10\)

Total number of random digits \(n\) = \(10000\)
\(x\) = \(931\)
\(\mu\) = \(np\)
\(\mu\) = \(10000\) X \(0.10\) = \(1000\)
\({\sigma}^2\) = \(np(1-p)\) = \(10000\) X \(0.1\) X \((1-0.1)\)
\({\sigma}^2\) = \(10000\) X \(0.1\) X \(0.9\) = \(900\)
\(\sigma\) = \(\sqrt{900}\) = \(30\)

The probability that the digit 3 appears at most \(931\) times is

\(P(S_{10000} \leqslant 931)\) = \(P(S^*_{10000} \leqslant \frac{x - \mu}{\sigma})\)

\(P(S_{10000} \leqslant 931)\) = \(P(S^*_{10000} \leqslant \frac{931 - 1000}{30})\)

\(P(S_{10000} \leqslant 931)\) = \(P(S^*_{10000} \leqslant -2.3)\) = \(0.0107\)

x <- 931
n <- 10000
p <- 1/10
mu <- n*p
sigmaSquare <- n*p*(1-p)
sigma <- sqrt(sigmaSquare)
S <- (x-mu)/sigma
v <- c(x, n, p, mu, sigmaSquare, sigma, S)
v

## [1]   931.0 10000.0     0.1  1000.0   900.0    30.0    -2.3

prob = round(pnorm(S), 4)
prob

## [1] 0.0107