Question

find the probability that in 200 tosses of a fair six - sided die, a five will be obtained at most 40 times.
a. 0.0853
b. 0.8810
c. 0.9131
d. 0.1190

Explanation:

Step1: Identify the binomial parameters

Let \(X\) be the number of fives. \(n = 200\), \(p=\frac{1}{6}\), \(q = 1 - p=\frac{5}{6}\).

Step2: Calculate the mean and standard deviation

Mean \(\mu=np=200\times\frac{1}{6}=\frac{100}{3}\approx33.33\).
Standard deviation \(\sigma=\sqrt{npq}=\sqrt{200\times\frac{1}{6}\times\frac{5}{6}}=\sqrt{\frac{500}{18}}\approx5.27\).

Step3: Apply the normal approximation (continuity correction)

We want \(P(X\leq40)\). Using continuity correction, we find \(P(X\leq40.5)\).
\(z=\frac{x-\mu}{\sigma}=\frac{40.5 - 33.33}{5.27}=\frac{7.17}{5.27}\approx1.36\).

Step4: Find the probability using the standard normal table

\(P(Z\leq1.36)\) from the standard - normal table is \(0.9131\).

Answer:

C. \(0.9131\)