Question

The length L in hours that a phone will work before it needs charge is normally distributed with a mean of 100 hours and a standard deviation of 15 hours.

i) Find p(L>127) and the 95^th percentile of L

II)Alice is about to go for a 6hour journey. Given that it is 127 hours since last Alice charged her phone .find the probability that her phone will not need charging before the journey is completed.

Answer 1

i) $L\sim N(\mu, \sigma^2)$

Given $\mu=100\ hr, \sigma=15\ hr$

$$P(L>127)=1-P(L\leq 127)$$

$$=1-P(L\leq \dfrac{127-100}{15})$$

$$=1-P(Z\leq1.8)\approx0.035930$$

$$P(Z<z)=0.95$$

$$z\approx1.644854$$

$$\dfrac{x-100}{15}\approx1.644854$$

$$x\approx124.67$$

ii)

$$127+6=133$$

$$P(L>133)=1-P(L\leq 133)$$

$$=1-P(L\leq \dfrac{133-100}{15})$$

$$=1-P(Z\leq2.2)\approx0.013903$$