In a bag of 345 chocolate candies, 36 of them are brown. The candy company claims that \( 13 \% \) of its plain chocolate candies are brown. For the following, assume that 7 of 7 . the claim of \( 13 \% \) is true, and assume that a sample consists of 345 chocolate candies. Complete parts (a) through (e) below. A. Yes, the result of 36 brown candies is less than the second value, so it is significantly low. \( 12.47 \%, 12 \) of 22 points B. No, the result of 36 brown candies is greater than the second value, so it is significantly high. D. Yes, the result of 36 brown candies lies between those limits, so it is neither significantly low nor significantly high. 36 brown candies is less than the first value, so it is significantly low. b. Find the probability of exactly 36 brown chocolate candies. The probability is \( \square \). (Round to four decimal places as needed.)

Alright, I need to tackle this probability problem step by step. Let's break down the information provided and understand what's being asked. We have a bag containing 345 chocolate candies, out of which 36 are brown. The candy company claims that 13% of its plain chocolate candies are brown. We're to assume that this claim is true and that a sample consists of 345 chocolate candies. The questions are: A. Determine whether the result of 36 brown candies is significantly low, high, or neither. B. Find the probability of exactly 36 brown chocolate candies. First, let's address part A. **Part A: Significance of 36 Brown Candies** Given: - Total candies: 345 - Number of brown candies: 36 - Claimed percentage of brown candies: 13% First, let's calculate the expected number of brown candies based on the company's claim. Expected number of brown candies = Total candies * Claimed percentage = 345 * 0.13 = 44.85 So, the expected number is approximately 44.85 brown candies. Now, we have 36 brown candies in the sample. To determine if this is significantly low, high, or neither, we can use a hypothesis test. Specifically, we'll perform a one-proportion z-test. **Hypotheses:** - Null hypothesis (\( H_0 \)): The proportion of brown candies is 13%. - Alternative hypothesis (\( H_1 \)): The proportion of brown candies is not 13%. **Calculating the Z-score:** First, find the standard error (SE) of the proportion. SE = \( \sqrt{\frac{p(1-p)}{n}} \) where \( p \) is the claimed proportion, and \( n \) is the sample size. SE = \( \sqrt{\frac{0.13(1-0.13)}{345}} \) SE ≈ \( \sqrt{\frac{0.13 \times 0.87}{345}} \) SE ≈ \( \sqrt{\frac{0.1131}{345}} \) SE ≈ \( \sqrt{0.000327} \) SE ≈ 0.0181 Now, calculate the Z-score. Z = \( \frac{\hat{p} - p}{SE} \) where \( \hat{p} \) is the observed proportion. Observed proportion \( \hat{p} = \frac{36}{345} \) \( \hat{p} ≈ 0.1043 \) Z = \( \frac{0.1043 - 0.13}{0.0181} \) Z ≈ \( \frac{-0.0257}{0.0181} \) Z ≈ -1.42 **Interpreting the Z-score:** A Z-score of -1.42 indicates that the observed proportion is 1.42 standard errors below the expected proportion. Using standard normal distribution tables or a calculator, a Z-score of -1.42 corresponds to a p-value of approximately 0.0786. **Decision Rule:** - If p-value < 0.05, reject \( H_0 \). - If p-value ≥ 0.05, fail to reject \( H_0 \). Since 0.0786 > 0.05, we fail to reject the null hypothesis. This means that the observed number of brown candies (36) is not significantly different from the expected number based on the company's claim. **Conclusion for Part A:** The result of 36 brown candies is neither significantly low nor significantly high. It lies within the expected range based on the company's claim. **Part B: Probability of Exactly 36 Brown Candies** To find the probability of exactly 36 brown candies, we'll use the binomial probability formula. The binomial probability formula is: P(X = k) = \( C(n, k) \times p^k \times (1-p)^{n-k} \) Where: - \( C(n, k) \) is the combination of n items taken k at a time. - \( p \) is the probability of success on a single trial. - \( n \) is the number of trials. - \( k \) is the number of successes. Given: - \( n = 345 \) - \( k = 36 \) - \( p = 0.13 \) First, calculate \( C(345, 36) \). Calculating combinations directly is computationally intensive, so we'll use the approximation: \( C(n, k) ≈ \frac{n^k}{k!} \) But since \( k \) is relatively large, it's better to use the exact combination formula or a calculator. Alternatively, we can use the normal approximation to the binomial distribution for large \( n \). **Normal Approximation:** Mean (\( \mu \)) = \( n \times p \) = 345 * 0.13 = 44.85 Variance (\( \sigma^2 \)) = \( n \times p \times (1-p) \) = 345 * 0.13 * 0.87 = 345 * 0.1131 = 38.9895 Standard deviation (\( \sigma \)) = \( \sqrt{38.9895} \) ≈ 6.24 Now, calculate the Z-score for 36 brown candies. Z = \( \frac{36 - 44.85}{6.24} \) Z ≈ \( \frac{-8.85}{6.24} \) Z ≈ -1.42 Using the standard normal distribution table, a Z-score of -1.42 corresponds to a cumulative probability of approximately 0.0786. However, since we're looking for the probability of exactly 36, we need to consider the continuity correction. Adjusted Z = \( \frac{36.5 - 44.85}{6.24} \) = \( \frac{-8.35}{6.24} \) ≈ -1.34 The cumulative probability for Z = -1.34 is approximately 0.0901. Therefore, the probability of exactly 36 brown candies is approximately 0.0901 or 9.01%. **Final Answer:** The probability of exactly 36 brown chocolate candies is approximately 0.0901.