Assume a random sample representative of the population is surveyed. If 36 people in the sample of size 117 are in favor, can the normal distribution approximation be used to estimate probabilities? (a) \( n p= \) (b) \( n(1-p)= \) (c) Yes, a normal distribution approximation can be used No, a normal distribution approximation can't be used

To determine whether the normal distribution approximation can be used to estimate probabilities for a sample proportion, we need to check two conditions: 1. \( n p \geq 10 \) 2. \( n (1 - p) \geq 10 \) Where: - \( n \) is the sample size. - \( p \) is the sample proportion of successes (in favor). Given: - Sample size \( n = 117 \) - Number of people in favor = 36 First, we calculate the sample proportion \( p \): \[ p = \frac{\text{Number of people in favor}}{n} = \frac{36}{117} \] Now, let's calculate \( n p \) and \( n (1 - p) \). ### Step 1: Calculate \( p \) \[ p = \frac{36}{117} \approx 0.3077 \] ### Step 2: Calculate \( n p \) \[ n p = 117 \times 0.3077 \approx 36 \] ### Step 3: Calculate \( n (1 - p) \) \[ 1 - p = 1 - 0.3077 \approx 0.6923 \] \[ n (1 - p) = 117 \times 0.6923 \approx 81 \] Now we can summarize the results: (a) \( n p \approx 36 \) (b) \( n(1 - p) \approx 81 \) ### Step 4: Check the conditions - \( n p = 36 \) (which is greater than 10) - \( n (1 - p) = 81 \) (which is also greater than 10) Since both conditions are satisfied, we can conclude: (c) Yes, a normal distribution approximation can be used.