The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours - Edexcel - A-Level Maths Statistics - Question 4 - 2013 - Paper 1

From the standard normal distribution, we need to find the Z-value where the cumulative probability is 0.10. This Z-value is approximately -1.2816.

Now, we convert this Z-score back to the original variable L:

$d = \mu + Z \cdot \sigma = 100 + (-1.2816) \cdot 15$

Calculating this gives:

$d = 100 - 19.224 = 80.776$

Thus, the value of d such that P(L < d) = 0.10 is approximately 80.776.

Since Alice's journey is for 6 hours and it has been 127 hours since her last charge, we need to find:

$P(L > 133)$

This can be calculated as:

$P(L > 133) = P(Z > 2.2)$

Standardizing as before gives:

$Z = \frac{133 - 100}{15} = \frac{33}{15} = 2.2$

Using the Z-table or calculator, we find:

$P(Z > 2.2) = 1 - P(Z < 2.2) = 1 - 0.9861 = 0.0139$

So the probability that Alice's phone will not need charging before her journey is approximately 0.0139.