Use Binomial Probability Formula: To solve part (a), we will use the binomial probability formula: $\newline$ $P(X = k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}$ $\newline$ where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single trial, and $\text{＂n choose k＂}$ is the binomial coefficient.
Calculate Binomial Coefficient: First, we calculate the binomial coefficient for $15$ choose $10$ : $\newline$ $\binom{15}{10} = \frac{15!}{10! \times (15-10)!}$
Calculate Probability of $10$ Successes: Now, we calculate the actual values: $\newline$ $(15 \choose 10) = \frac{15!}{10! \times 5!} = \frac{(15 \times 14 \times 13 \times 12 \times 11)}{(5 \times 4 \times 3 \times 2 \times 1)} = 3003$
Calculate Mean and Standard Deviation: Next, we calculate the probability of exactly $10$ successes: $\newline$ $P(X = 10) = 3003 \times (0.55)^{10} \times (0.45)^5$
Apply Continuity Correction Factor: Performing the calculation: $\newline$ $P(X = 10) = 3003 \times 0.000025937 \times 0.01849 \approx 0.142$
Convert Values to Z-Scores: Round the answer to three decimal places as instructed: $\newline$ $P(X = 10) \approx 0.142$
Find Probabilities from Z-Scores: For part (b), we will use the normal approximation to the binomial distribution. The mean $\mu$ and standard deviation $\sigma$ of the binomial distribution can be calculated using the formulas: $\newline$ $\mu = n \times p$ $\newline$ $\sigma = \sqrt{n \times p \times (1 - p)}$
Calculate Probability of $10$ Successes: Calculate the mean ( $\mu$ ): $\mu = 15 \times 0.55 = 8.25$
Compare Results of Parts: Calculate the standard deviation ( $\sigma$ ):\sigma = \sqrt{$15$ \times $0$.$55$ \times $0$.$45$} \approx \sqrt{$3$.$7125$} \approx \(1. $927$
Compare Results of Parts: Calculate the standard deviation ( $\sigma$ ): $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ To find the probability of exactly $10$ successes, we need to use the continuity correction factor. We will find the probability that the number of successes is between $9.5$ and $10.5$ .
Compare Results of Parts: Calculate the standard deviation ( $\sigma$ ): $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ To find the probability of exactly $10$ successes, we need to use the continuity correction factor. We will find the probability that the number of successes is between $9$ . $5$ and $10$ . $5$ .We convert these values to z-scores using the formula: $z = (x - \mu) / \sigma$
Compare Results of Parts: Calculate the standard deviation ( $\sigma$ ): $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ To find the probability of exactly $10$ successes, we need to use the continuity correction factor. We will find the probability that the number of successes is between $9$ . $5$ and $10$ . $5$ .We convert these values to z-scores using the formula: $z = \frac{x - \mu}{\sigma}$ Calculate the z-scores for $9$ . $5$ and $10$ . $5$ : $z(9.5) = \frac{9.5 - 8.25}{1.927} \approx 0.649$ $z(10.5) = \frac{10.5 - 8.25}{1.927} \approx 1.169$
Compare Results of Parts: Calculate the standard deviation ( $\sigma$ ): $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ To find the probability of exactly $10$ successes, we need to use the continuity correction factor. We will find the probability that the number of successes is between $9.5$ and $10.5$ .We convert these values to z-scores using the formula: $z = (x - \mu) / \sigma$ Calculate the z-scores for $9.5$ and $10.5$ : $z(9.5) = (9.5 - 8.25) / 1.927 \approx 0.649$ $z(10.5) = (10.5 - 8.25) / 1.927 \approx 1.169$ Now, we look up these z-scores in the standard normal distribution table or use a calculator to find the probabilities: $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ $0$ $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ $1$
Compare Results of Parts: Calculate the standard deviation ( $\sigma$ ): $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ To find the probability of exactly $10$ successes, we need to use the continuity correction factor. We will find the probability that the number of successes is between $9$ . $5$ and $10$ . $5$ .We convert these values to z-scores using the formula: $z = \frac{x - \mu}{\sigma}$ Calculate the z-scores for $9$ . $5$ and $10$ . $5$ : $z(9.5) = \frac{9.5 - 8.25}{1.927} \approx 0.649$ $z(10.5) = \frac{10.5 - 8.25}{1.927} \approx 1.169$ Now, we look up these z-scores in the standard normal distribution table or use a calculator to find the probabilities: $P(z < 0.649) \approx 0.7422$ $P(z < 1.169) \approx 0.8790$ The probability of exactly $10$ successes is approximately the difference between these two probabilities: $P(9.5 < X < 10.5) \approx P(z < 1.169) - P(z < 0.649) \approx 0.8790 - 0.7422 \approx 0.1368$
Compare Results of Parts: Calculate the standard deviation ( $\sigma$ ): $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ To find the probability of exactly $10$ successes, we need to use the continuity correction factor. We will find the probability that the number of successes is between $9$ . $5$ and $10$ . $5$ .We convert these values to z-scores using the formula: $z = \frac{x - \mu}{\sigma}$ Calculate the z-scores for $9$ . $5$ and $10$ . $5$ : $z(9.5) = \frac{9.5 - 8.25}{1.927} \approx 0.649$ $z(10.5) = \frac{10.5 - 8.25}{1.927} \approx 1.169$ Now, we look up these z-scores in the standard normal distribution table or use a calculator to find the probabilities: $P(z < 0.649) \approx 0.7422$ $P(z < 1.169) \approx 0.8790$ The probability of exactly $10$ successes is approximately the difference between these two probabilities: $P(9.5 < X < 10.5) \approx P(z < 1.169) - P(z < 0.649) \approx 0.8790 - 0.7422 \approx 0.1368$ Round the answer to four decimal places as instructed: $P(9.5 < X < 10.5) \approx 0.1368$
Compare Results of Parts: Calculate the standard deviation ( $\sigma$ ): $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ To find the probability of exactly $10$ successes, we need to use the continuity correction factor. We will find the probability that the number of successes is between $9$ . $5$ and $10$ . $5$ .We convert these values to z-scores using the formula: $z = (x - \mu) / \sigma$ Calculate the z-scores for $9$ . $5$ and $10$ . $5$ : $z(9.5) = (9.5 - 8.25) / 1.927 \approx 0.649$ $z(10.5) = (10.5 - 8.25) / 1.927 \approx 1.169$ Now, we look up these z-scores in the standard normal distribution table or use a calculator to find the probabilities: $P(z < 0.649) \approx 0.7422$ $P(z < 1.169) \approx 0.8790$ The probability of exactly $10$ successes is approximately the difference between these two probabilities: $P(9.5 < X < 10.5) \approx P(z < 1.169) - P(z < 0.649) \approx 0.8790 - 0.7422 \approx 0.1368$ Round the answer to four decimal places as instructed: $P(9.5 < X < 10.5) \approx 0.1368$ For part (c), we compare the results of parts (a) and (b). The binomial probability of exactly $10$ successes is $0.142$ , and the normal approximation gives us $\sigma = \sqrt{15 \times 0.55 \times 0.45} \approx \sqrt{3.7125} \approx 1.927$ $0$ . These results are fairly close, but not exactly the same.