A packing plant fills bags with cement - Edexcel - A-Level Maths Statistics - Question 7 - 2008 - Paper 2

To find the weight that is exceeded by 99% of the bags, we need to find the Z-score that corresponds to the 1% in the upper tail:

From Z-tables, the Z-score that corresponds to 0.01 is approximately -2.326.

Now we can convert this Z-score back to the weight (X) using the formula:

$X = \mu + Z \cdot \sigma$

Substituting our values:

$X = 50 + (-2.326) \cdot 2 \approx 45.348$

Thus, 99% of the bags weigh less than approximately 45.35 kg.

Let (p = P(X > 53) \approx 0.0668) and thus (q = P(X < 53) = 1 - p \approx 0.9332).

The scenario involves selecting 3 bags: 2 weigh more than 53 kg and 1 weighs less than 53 kg.

Using the binomial probability formula:

$P(k; n, p) = \binom{n}{k} p^k q^{n-k}$

Here, (n=3), (k=2), (p=0.0668), (q=0.9332):

$P(2 \text{ more than } 53 ext{ and } 1 \text{ less than } 53) = \binom{3}{2} (0.0668)^2 (0.9332)^1$

Calculating:

$= 3 \cdot (0.0668)^2 \cdot (0.9332) \approx 0.012424$

Therefore, the probability is approximately 0.0124.