Discussion 9

Exercise - 1, Page - 338, Section - 9.1

\((1)\) Let \(S_100\) be the number of heads that turn up in 100 tosses of a fair coin. Use the Central Limit Theorem to estimate

\((a)\) P(\(S_100\) <= 45)

Answer

Note: refer Z-table for z-values

n = 100
p = \(\frac{1}{2}\)
q = \(\frac{1}{2}\)

P(\(S_100\) <= 45)

\(\sqrt{npq} = \sqrt{100 * 1/2 * 1/2} = 5\) np = 100 * 1/2 = 50
\[P(S_100 <= 45) = P(S_n <= \frac{45 + \frac {1}{2}- 50} {5} = \frac{45 + \frac {1}{2} - 50}{5} = \frac{-5.5}{5} = -1.1) = 0.1356661\]

# using R
pnorm(-1.1)

## [1] 0.1356661

\((b)\) P(45 < S100 < 55)

Answer

\[P(45 < S_100 < 55) = P(45 < S_n < 55) = \frac{45 + \frac {1}{2} - 50} {5} < S_n < \frac{55 + \frac {1}{2} - 50}{5} \] \[P(45 < S_n < 55) = \frac {-5.5} {5} < S_n < \frac {5.5} {5}\] \[P(45 < S_n < 55) = -1.1 < S_n < 1.1 = NA(-1.1,1.1) = 2NA(0,1.1) = 2 * 0.3643 = 0.7286\]

# using R
pnorm(1.1) - pnorm(-1.1)

## [1] 0.7286679

\((c)\) P(S100 > 63)

Answer

\[P(S_100 > 63) = 1 - P( S_n < 63) = 1 - S_n < \frac{63 + \frac {1}{2} - 50}{5} = 1 - S_n < (\frac {13.5} {5}) \]

\[P(S_100 > S_n) = 1 - S_n (2.7) = 1 - NA(2.7) = 1 - 0.99653 = 0.00347\]

# using R
1 - pnorm(2.7)

## [1] 0.003466974

\((d)\) P(S100 < 57)

Answer

\[P(S_100 < 57) = P(S_n < \frac{57 + \frac {1}{2} - 50} {5}) = \frac{57.5 - 50}{5}\] \[P(S_100 < 57) = \frac{7.5}{5} = 1.5 = NA(1.5) = 0.93319\]

# using R
pnorm(1.5)

## [1] 0.9331928