The weights of bags of popcorn are normally distributed with mean of 200 g and 60% of all bags weighing between 190 g and 210 g - Edexcel - A-Level Maths Statistics - Question 6 - 2008 - Paper 1

Given that 60% of bags weigh between 190 g and 210 g, we can standardize these values. The z-scores corresponding to these weights can be calculated as:

For 190 g: $Z = \frac{X - \mu}{\sigma} = \frac{190 - 200}{\sigma}$

For 210 g: $Z = \frac{210 - 200}{\sigma}$

Since 60% of the population falls between these two z-scores, we find: $P(Z < z_2) - P(Z < z_1) = 0.60$

Using standard normal distribution tables, we find the z-scores corresponding to 0.3 (0.6 is half on both sides). Solving gives: $Z = \frac{10}{\sigma}$.

From the tables, we find z values, allowing us to compute: $\sigma \approx 11.88 g$.

To find the probability that a customer will complain, we calculate: $P(X < 180) = P\left(Z < \frac{180 - 200}{\sigma}\right) = P\left(Z < \frac{-20}{11.88}\right) \approx P(Z < -1.683)$

Using standard normal distribution tables:

$P(Z < -1.683) \approx 0.046$

Thus, the probability that a customer will complain is approximately 0.046, or 4.6%.