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

To find the standard deviation, we can use the information given about the percentage of bags that weigh between 190 g and 210 g. Since 60% of the bags fall within this range, we are looking at a range of ±10 g from the mean (200 g).

Using the Z-scores:

• For the lower bound (190 g), we can find the Z-score as: $Z = \frac{X - \mu}{\sigma} = \frac{190 - 200}{\sigma}$
• For the upper bound (210 g): $Z = \frac{210 - 200}{\sigma} = \frac{10}{\sigma}$

Since the probability between these Z-scores corresponds to 60%, which translates to Z-scores of approximately -0.8416 and +0.8416, we can set up the equations:

Using the properties of the normal distribution: $P(Z < 0.8416) - P(Z < -0.8416) = 0.6$ This information allows us to calculate the standard deviation as: $\sigma \approx 11.882129$

To find the probability that a customer complains (i.e., their bag weighs less than 180 g), we use the Z-score:

$Z = \frac{180 - 200}{\sigma} = \frac{-20}{11.882129} \approx -1.6832$

Now, we determine the probability: $P(X < 180) = P(Z < -1.6832)$ Using the standard normal distribution table, we find: (P(Z < -1.6832) \approx 0.0465) Thus, the probability that a customer will complain is approximately 0.0465, or 4.65%.