A phone manufacturer models the battery life, T hours, of a fully charged phone under normal use by T N(11.5, 1.2^2). (a) Find P(T>13). [3] (b) Find P(10<T<12.7). [4]

Answer: ((a)) P(T>13)=0.1056 ((b)) P(10 13) z=13-11.5/1.2=1.5/1.2=1.25 P(T>13)=P(Z>1.25)=1- (1.25) From the table, (1.25)=0.8944, so P(T>13)=1-0.8944=0.1056 Part (b): P(10 < T <…

A Level Maths 9709: Probabilities from a Normal Distribution

Setting up

$T\sim N(11.5,1.2^2)$ , so $\mu=11.5$ and $\sigma=1.2$ . Every probability is found via $Z=\dfrac{T-\mu}{\sigma}$ .

Part (a): P(T > 13)

$z=\frac{13-11.5}{1.2}=\frac{1.5}{1.2}=1.25$

$P(T>13)=P(Z>1.25)=1-\Phi(1.25)$

From the table, $\Phi(1.25)=0.8944$ , so $P(T>13)=1-0.8944=0.1056$

Part (b): P(10 < T < 12.7)

Standardise both boundaries.

Lower boundary: $z_1=\frac{10-11.5}{1.2}=\frac{-1.5}{1.2}=-1.25$

Upper boundary: $z_2=\frac{12.7-11.5}{1.2}=\frac{1.2}{1.2}=1$

$P(10<T<12.7)=P(-1.25<Z<1)=\Phi(1)-\Phi(-1.25)$

Using symmetry, $\Phi(-1.25)=1-\Phi(1.25)=1-0.8944=0.1056$ , and $\Phi(1)=0.8413$ from the table.

$P(10<T<12.7)=0.8413-0.1056=0.7357$

Check: the interval has width $12.7-10=2.7$ hours $=2.25\sigma$ , split unevenly around the mean ( $1$ hour below the mean, $1.2$ hours above it), so leaving out roughly a quarter of the total probability across both tails combined is a sensible order of magnitude.

Final answers

(a) $P(T>13)=\boxed{0.1056}$
(b) $P(10<T<12.7)=\boxed{0.7357}$