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 use the information that 60% of the bags weigh between 190 g and 210 g. We express this as:

$P(190 < X < 210) = 0.6$

With a mean ($\mu$) of 200 g, we convert these bounds to z-scores:

For X = 190: $z_1 = \frac{190 - 200}{\sigma} = \frac{-10}{\sigma}$

For X = 210: $z_2 = \frac{210 - 200}{\sigma} = \frac{10}{\sigma}$

The probability corresponds to: $P(z_1 < Z < z_2) = 0.6$

Using standard normal distribution tables, we find the z-values that correspond to this cumulative probability. Given that the area in between those z-values is 0.6, we can work out that:

• The area to the left of $z_1$ is 0.2 (or $P(Z < z_1)$).
• The area to the left of $z_2$ is 0.8 (or $P(Z < z_2)$).

Thus, we conclude:

• $z_1$ corresponds approximately to -0.8416 and $z_2$ corresponds to 0.8416.

Using: rac{10}{\sigma} = 0.8416

We find: $\sigma \approx \frac{10}{0.8416} \approx 11.88 g$

Customers will complain if their bag weighs less than 180 g. We calculate:

$P(X < 180) = P\left(Z < \frac{180 - 200}{\sigma}\right)$ Substituting the value of $\sigma$: $= P\left(Z < \frac{-20}{11.88}\right) \approx P(Z < -1.6832)$ Using the standard normal tables, we find: $P(Z < -1.6832) \approx 0.0465$ Thus, the probability that a customer will complain is approximately 4.65%.