b) A major department store has determined that its customers charge an average of \( N \$ 500 \) per month, with a standard deviation of \( N \$ 80 \). Assume the amounts of charges are normally distributed. (i) What percentage of customers charges more than \( N \$ 380 \) per month? (ii) What percentage of customers charges less than \( N \$ 340 \) per month? (iii) What percentage of customers charges between \( N \$ 644 \) and \( N \$ 700 \) per month?

To solve these problems, we will use the properties of the normal distribution. Given that the average charge is \( \mu = 500 \) and the standard deviation is \( \sigma = 80 \), we can standardize the values using the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( X \) is the value we are interested in. ### (i) What percentage of customers charges more than \( N \$ 380 \) per month? 1. Calculate the z-score for \( N \$ 380 \): \[ z = \frac{(380 - 500)}{80} = \frac{-120}{80} = -1.5 \] 2. Look up the z-score of \(-1.5\) in the standard normal distribution table, or use a calculator. The cumulative probability for \( z = -1.5 \) is approximately \( 0.0668 \). 3. To find the percentage of customers who charge more than \( N \$ 380 \): \[ P(X > 380) = 1 - P(X \leq 380) = 1 - 0.0668 = 0.9332 \] 4. Convert to percentage: \[ 0.9332 \times 100 \approx 93.32\% \] ### (ii) What percentage of customers charges less than \( N \$ 340 \) per month? 1. Calculate the z-score for \( N \$ 340 \): \[ z = \frac{(340 - 500)}{80} = \frac{-160}{80} = -2.0 \] 2. Look up the z-score of \(-2.0\) in the standard normal distribution table. The cumulative probability for \( z = -2.0 \) is approximately \( 0.0228 \). 3. Therefore, the percentage of customers who charge less than \( N \$ 340 \): \[ P(X < 340) \approx 0.0228 \] 4. Convert to percentage: \[ 0.0228 \times 100 \approx 2.28\% \] ### (iii) What percentage of customers charges between \( N \$ 644 \) and \( N \$ 700 \) per month? 1. Calculate the z-scores for \( N \$ 644 \) and \( N \$ 700 \): For \( N \$ 644 \): \[ z_{644} = \frac{(644 - 500)}{80} = \frac{144}{80} = 1.8 \] For \( N \$ 700 \): \[ z_{700} = \frac{(700 - 500)}{80} = \frac{200}{80} = 2.5 \] 2. Look up the z-scores in the standard normal distribution table: - For \( z = 1.8 \), the cumulative probability is approximately \( 0.9641 \). - For \( z = 2.5 \), the cumulative probability is approximately \( 0.9938 \). 3. To find the percentage of customers who charge between \( N \$ 644 \) and \( N \$ 700 \): \[ P(644 < X < 700) = P(X < 700) - P(X < 644) = 0.9938 - 0.9641 = 0.0297 \] 4. Convert to percentage: \[ 0.0297 \times 100 \approx 2.97\% \] ### Summary of Results: - (i) Approximately **93.32%** of customers charge more than \( N \$ 380 \). - (ii) Approximately **2.28%** of customers charge less than \( N \$ 340 \). - (iii) Approximately **2.97%** of customers charge between \( N \$ 644 \) and \( N \$ 700 \).