A manufacturer fills jars with coffee - Edexcel - A-Level Maths Statistics - Question 7 - 2012 - Paper 1

To find the value of w, we need to determine the Z-value that corresponds to the cumulative probability of 0.80 (since P(232 < W < w) implies we are looking at the upper tail):

Using the Z-table, we find:

$P(Z < z) = 0.80 \Rightarrow z \approx 0.84$

Using standardization, we set up the equation:

$z = \frac{w - 232}{5} = 0.84$

Solving for w:

\Rightarrow w = 232 + 4.2 = 236.2$$ Hence, $$w \approx 236.2$$.

Let p be the probability that a single jar contains between 232 and w grams:

$p = P(232 < W < w) = 0.20$
This implies the probability of not being within this range is:

$q = 1 - p = 0.80$

The scenario of only one jar containing between 232 grams and w grams can occur in two ways:

1. The first jar is within the range, and the second is not.
2. The first jar is not within the range, and the second is.

Thus, the total probability is:

$P(\text{only one}) = 2 \times p \times q = 2 \times 0.20 \times 0.80 = 0.32$

So, the probability that only one jar contains between 232 grams and w grams of coffee is:

$P(\text{only one}) \approx 0.32$.