Question

A laptop company claims up to 8.1 hours of wireless web usage for its newest laptop battery life. However, reviews on this laptop shows many complaints about low battery life. A survey on battery life reported by customers shows that it follows a normal distribution with mean 7.5 hours and standard deviation 36 minutes.

(a) What is the probability that the battery life is at least 8.1 hours?
Round your answer to four decimal places (e.g. 98.7654).

(b) What is the probability that the battery life is less than 6.9 hours?
Round your answer to four decimal places (e.g. 98.7654).

(c) What is the time of use that is exceeded with probability 0.9?
Round your answer to two decimal places (e.g. 98.76).

Answer 1

Let X = the random variable denoting the laptop battery life

We are given, X ~ N($\mu$ = 7.5, $\sigma^2=(\frac{36}{60})^2$= 0.6²)

Then by the definition of standard normal distribution we have,

Z = $\frac{X-\mu}{\sigma}$ ~ N(0, 1)

(a) The probability that the battery life is at least 8.1 hours

= P(X $\geq$ 8.1)

= P($\frac{X-7.5}{0.6}\geq\frac{8.1-7.5}{0.6}$)

= P(Z $\geq$ 1)

= 1 - P(Z < 1)

= 1 - $\Phi$(1)

= 1 - 0.8413 [using standard normal distribution table]

= 0.1587

Answer: The probability that the battery life is at least 8.1 hours is 0.1587.

(b) The probability that the battery life is less than 6.9 hours

= P(X < 6.9)

= P($\frac{X-7.5}{0.6}<\frac{6.9-7.5}{0.6}$)

= P(Z < - 1)

= $\Phi$(- 1)

= 0.1587

Answer: The probability that the battery life is less than 6.9 hours is 0.1587.

(c) Let t be the time of use that is exceeded with probability 0.9

i.e. P(X > t) = 0.9

i.e. P($\frac{X-7.5}{0.6}>\frac{t-7.5}{0.6}$) = 0.9

i.e. P(Z > $\frac{t-7.5}{0.6}$) = 0.9

i.e. 1 - P(Z $\leq\frac{t-7.5}{0.6}$) = 0.9

i.e. P(Z $\leq\frac{t-7.5}{0.6}$ )= 1 - 0.9 = 0.1

i.e. $\Phi(\frac{t-7.5}{0.6}$) = $\Phi$(- 1.28)

i.e. $\frac{t-7.5}{0.6}$ = - 1.28

i.e. t = 6.73 (rounded to 2 decimal places)

Answer: The time of use that is exceeded with probability 0.9 is 6.73 hours.