Question

find the probability that in 200 tosses of a fair six - sided die, a five will be obtained at least 40 times.
a. 0.3875
b. 0.1210
c. 0.8910
d. 0.0871

Explanation:

Step1: Identify the distribution parameters

This is a binomial distribution problem. Let \(n = 200\) (number of trials), \(p=\frac{1}{6}\) (probability of getting a five in a single toss of a fair six - sided die), and \(q = 1 - p=\frac{5}{6}\).

The mean of the binomial distribution is \(\mu=np=200\times\frac{1}{6}=\frac{100}{3}\approx33.33\)

The standard deviation is \(\sigma=\sqrt{npq}=\sqrt{200\times\frac{1}{6}\times\frac{5}{6}}=\sqrt{\frac{500}{18}}\approx5.27\)

Step2: Apply the normal approximation to the binomial

We use the normal approximation \(X\sim N(np, npq)\). For the binomial - to - normal approximation, we apply the continuity correction. To find \(P(X\geq40)\), we find \(P(X > 39.5)\) when using the normal approximation.

We calculate the z - score: \(z=\frac{x-\mu}{\sigma}\), where \(x = 39.5\), \(\mu=\frac{100}{3}\approx33.33\), and \(\sigma\approx5.27\)

\(z=\frac{39.5 - 33.33}{5.27}=\frac{6.17}{5.27}\approx1.17\)

Step3: Find the probability using the standard normal table

\(P(Z\geq1.17)=1 - P(Z < 1.17)\)

From the standard normal table, \(P(Z < 1.17)=0.8790\)

\(P(Z\geq1.17)=1 - 0.8790 = 0.1210\)

Answer:

B. \(0.1210\)