QUESTION 1 a sample of 100 school children who were given two toothpaste A and B, which were identical apart from the flavor were asked which they preferred. Of the 100 children, 65 said they preferred A. If in a general population of children, A and B are equally preferred, what is the probability that in a sample of 100 children, 65 or more prefer A?

To determine the probability that **65 or more** out of **100** children prefer toothpaste **A** when, in the general population, both toothpastes are equally preferred (\( p = 0.5 \)), we can model this scenario using the **Binomial Distribution**. ### Step 1: Define the Parameters - **Number of trials (n):** 100 - **Probability of success (p):** 0.5 (since A and B are equally preferred) ### Step 2: Calculate the Mean and Standard Deviation - **Mean (\( \mu \)):** \( n \times p = 100 \times 0.5 = 50 \) - **Standard Deviation (\( \sigma \)):** \( \sqrt{n \times p \times (1 - p)} = \sqrt{100 \times 0.5 \times 0.5} = 5 \) ### Step 3: Use Normal Approximation Since \( n \) is large and \( p \) is neither very close to 0 nor 1, the **Normal Approximation** to the Binomial Distribution is appropriate. - **Continuity Correction:** To find \( P(X \geq 65) \), we use \( P(X \geq 64.5) \) for the normal approximation. - **Z-score Calculation:** \[ Z = \frac{64.5 - \mu}{\sigma} = \frac{64.5 - 50}{5} = 2.9 \] ### Step 4: Find the Probability Using the Z-score Using standard normal distribution tables or a calculator: - **\( P(Z \geq 2.9) \):** Approximately **0.0018** or **0.18%** ### Conclusion There is approximately a **0.18%** probability that **65 or more** out of **100** children would prefer toothpaste **A** under the assumption that both toothpastes are equally preferred in the general population. **Answer:** There is approximately a 0.18 % probability of observing 65 or more children preferring A.