A biased coin has probability \(p = 0.6\) of landing heads.
The coin is tossed \(n = 100\) times.
Find \(P(X \ge 65)\) using a normal approximation.
Step 1: Check conditions
\[
np = 100(0.6) = 60
\]
\[
n(1-p) = 100(0.4) = 40
\]
Both are greater than 5, so the approximation is valid.
Step 2: Mean and variance
\[
\mu = np = 60
\]
\[
\sigma^2 = np(1-p) = 100(0.6)(0.4) = 24
\]
\[
\sigma = \sqrt{24} \approx 4.90
\]
Step 3: Continuity correction
\[
P(X \ge 65) \approx P(X > 64.5)
\]
Step 4: Standardise
\[
Z = \frac{64.5 - 60}{4.90}
\]
\[
Z \approx 0.92
\]
Step 5: Use normal tables
\[
P(Z > 0.92) = 0.1788
\]
Final answer:
\[
P(X \ge 65) \approx 0.179
\]