Questions: Random samples of size n=95 were selected from a binomial population with p=0.2. Use the normal distribution to approximate the following probability. (Round your answer to four decimal places) P(β>0.19)=

Solution Steps

For a binomial distribution with parameters \( n = 95 \) and \( p = 0.2 \), we calculate the mean \( \mu \) and standard deviation \( \sigma \) as follows:

\[ \mu = n \cdot p = 95 \cdot 0.2 = 19.0 \]

The variance \( \sigma^2 \) is given by:

\[ \sigma^2 = n \cdot p \cdot q = 95 \cdot 0.2 \cdot (1 - 0.2) = 95 \cdot 0.2 \cdot 0.8 = 15.2 \]

Thus, the standard deviation \( \sigma \) is:

\[ \sigma = \sqrt{15.2} \approx 3.8987 \]

To find \( P(\beta > 0.19) \), we first convert the sample proportion to the number of successes:

\[ x = \beta \cdot n = 0.19 \cdot 95 = 18.05 \]

Next, we need to find the Z-scores for the normal approximation. The Z-score for \( x \) is calculated as:

\[ Z = \frac{x - \mu}{\sigma} = \frac{18.05 - 19.0}{3.8987} \approx -0.2437 \]

Using the properties of the standard normal distribution, we find:

\[ P(\beta > 0.19) = P(X > 18.05) = 1 - P(Z < -0.2437) = \Phi(\infty) - \Phi(-0.2437) \]

From the standard normal distribution table, we find:

\[ \Phi(-0.2437) \approx 0.4037 \]

Thus, the probability is:

\[ P(\beta > 0.19) = 1 - 0.4037 = 0.5963 \]

Final Answer