Hypoglycemia is the most common serious side effect of insulin. It occurs in approximately \( 16 \% \) of Type I diabetic patie Consider a sample of 100 Type I diabetics taking insulin. Round to 4 decimal places (a) \( P \) (more than 13 having hypoglycemia \( )= \) (b) \( P(12 \) to 33 patients having hypoglycemia \( )= \)

To solve this problem, we can model the number of Type I diabetics experiencing hypoglycemia as a binomial distribution. The parameters for this distribution are: - \( n = 100 \) (the number of trials, or patients) - \( p = 0.16 \) (the probability of success, or the probability that a patient experiences hypoglycemia) We can use the normal approximation to the binomial distribution since \( n \) is large. The mean \( \mu \) and standard deviation \( \sigma \) of the binomial distribution can be calculated as follows: \[ \mu = n \cdot p \] \[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} \] Let's calculate \( \mu \) and \( \sigma \) first. Then we will use the normal approximation to find the probabilities for parts (a) and (b). ### Step 1: Calculate \( \mu \) and \( \sigma \) \[ \mu = 100 \cdot 0.16 \] \[ \sigma = \sqrt{100 \cdot 0.16 \cdot (1 - 0.16)} \] ### Step 2: Calculate the probabilities #### (a) \( P \) (more than 13 having hypoglycemia) To find \( P(X > 13) \), we can use the normal approximation: \[ P(X > 13) = 1 - P(X \leq 13) \] Using the continuity correction, we will calculate \( P(X \leq 13.5) \). #### (b) \( P(12 \) to 33 patients having hypoglycemia) To find \( P(12 \leq X \leq 33) \), we will calculate: \[ P(12 \leq X \leq 33) = P(X \leq 33) - P(X < 12) \] Using the continuity correction, we will calculate \( P(X \leq 33.5) \) and \( P(X < 11.5) \). Now, let's perform the calculations for \( \mu \) and \( \sigma \), and then find the probabilities. Calculate the value by following steps: - step0: Calculate: \(100\times 0.16\) - step1: Multiply the numbers: \(16\) Calculate or simplify the expression \( \sqrt(100 * 0.16 * (1 - 0.16)) \). Calculate the value by following steps: - step0: Calculate: \(\sqrt{100\times 0.16\left(1-0.16\right)}\) - step1: Subtract the numbers: \(\sqrt{100\times 0.16\times 0.84}\) - step2: Multiply the terms: \(\sqrt{13.44}\) - step3: Rewrite the expression: \(\sqrt{\frac{336}{25}}\) - step4: Use the properties of radicals: \(\frac{\sqrt{336}}{\sqrt{25}}\) - step5: Simplify the expression: \(\frac{4\sqrt{21}}{5}\) The calculations yield the following results: - Mean \( \mu = 16 \) - Standard deviation \( \sigma \approx 3.6661 \) ### Step 3: Calculate the probabilities #### (a) \( P(X > 13) \) Using the continuity correction, we need to find \( P(X \leq 13.5) \). To standardize this, we use the z-score formula: \[ z = \frac{x - \mu}{\sigma} \] For \( x = 13.5 \): \[ z = \frac{13.5 - 16}{3.6661} \] #### (b) \( P(12 \leq X \leq 33) \) We need to calculate: 1. \( P(X \leq 33.5) \) 2. \( P(X < 11.5) \) For \( x = 33.5 \): \[ z_{33.5} = \frac{33.5 - 16}{3.6661} \] For \( x = 11.5 \): \[ z_{11.5} = \frac{11.5 - 16}{3.6661} \] Now, let's calculate the z-scores for both parts. Calculate the value by following steps: - step0: Calculate: \(\frac{\left(33.5-16\right)}{3.6661}\) - step1: Remove the parentheses: \(\frac{33.5-16}{3.6661}\) - step2: Subtract the numbers: \(\frac{17.5}{3.6661}\) - step3: Convert the expressions: \(\frac{\frac{35}{2}}{\frac{36661}{10000}}\) - step4: Multiply by the reciprocal: \(\frac{35}{2}\times \frac{10000}{36661}\) - step5: Reduce the numbers: \(35\times \frac{5000}{36661}\) - step6: Multiply: \(\frac{35\times 5000}{36661}\) - step7: Multiply: \(\frac{175000}{36661}\) Calculate or simplify the expression \( (11.5 - 16) / 3.6661 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(11.5-16\right)}{3.6661}\) - step1: Remove the parentheses: \(\frac{11.5-16}{3.6661}\) - step2: Subtract the numbers: \(\frac{-4.5}{3.6661}\) - step3: Convert the expressions: \(\frac{-\frac{9}{2}}{\frac{36661}{10000}}\) - step4: Multiply by the reciprocal: \(-\frac{9}{2}\times \frac{10000}{36661}\) - step5: Reduce the numbers: \(-9\times \frac{5000}{36661}\) - step6: Multiply: \(-\frac{9\times 5000}{36661}\) - step7: Multiply: \(-\frac{45000}{36661}\) Calculate or simplify the expression \( (13.5 - 16) / 3.6661 \). Calculate the value by following steps: - step0: Calculate: \(\frac{\left(13.5-16\right)}{3.6661}\) - step1: Remove the parentheses: \(\frac{13.5-16}{3.6661}\) - step2: Subtract the numbers: \(\frac{-2.5}{3.6661}\) - step3: Convert the expressions: \(\frac{-\frac{5}{2}}{\frac{36661}{10000}}\) - step4: Multiply by the reciprocal: \(-\frac{5}{2}\times \frac{10000}{36661}\) - step5: Reduce the numbers: \(-5\times \frac{5000}{36661}\) - step6: Multiply: \(-\frac{5\times 5000}{36661}\) - step7: Multiply: \(-\frac{25000}{36661}\) It seems that I attempted to use the wrong tool for finding the probabilities from the z-scores. Instead, I will provide the approximate probabilities based on standard normal distribution values. Using a standard normal distribution table or calculator: 1. **For \( z_{13.5} \approx -0.6819 \)**: - \( P(Z \leq -0.6819) \approx 0.2483 \) 2. **For \( z_{11.5} \approx -1.2275 \)**: - \( P(Z \leq -1.2275) \approx 0.1093 \) Now we can calculate the probabilities for parts (a) and (b): ### Final Calculations #### (a) \( P(X > 13) \) \[ P(X > 13) = 1 - P(X \leq 13.5) \approx 1 - 0.2483 = 0.7517 \] #### (b) \( P(12 \leq X \leq 33) \) \[ P(12 \leq X \leq 33) = P(X \leq 33.5) - P(X < 11.5) \approx 1 - 0.1093 = 0.8907 \] ### Summary of Results - (a) \( P(X > 13) \approx 0.7517 \) - (b) \( P(12 \leq X \leq 33) \approx 0.8907 \) Both probabilities are rounded to four decimal places.