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

To find the value of d for which P(L < d) = 0.10, we first need the Z-value that corresponds to this probability. From Z-tables, we find:

$P(Z < d) \Rightarrow Z \approx -1.2816$

Now, we convert this Z-value back to L:

$L = \mu + Z \cdot \sigma \Rightarrow d = 100 + (-1.2816) \cdot 15 \approx 80.76$

Alice's journey lasts 6 hours, and she has already used the phone for 127 hours. Thus, the total time until the journey finishes is:

$L < 127 + 6 = 133$

We find the probability:

$P(L > 133) = P(L < 133) = 1 - P(Z < 2.2)$

Standardizing again:

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

From Z-tables:

$P(Z < 2.2) \approx 0.9861$

Thus:

$P(L > 133) \approx 1 - 0.9861 = 0.0139$