Question

A laptop company claims up to 9.0 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 8.5 hours and standard deviation 39 minutes.

(a) What is the probability that the battery life is at least 9.0 hours?

Round your answer to four decimal places (e.g. 98.7654).

P= Enter your answer in accordance to the item a) of the question statement

(b) What is the probability that the battery life is less than 8.0 hours?

Round your answer to four decimal places (e.g. 98.7654).

P= Enter your answer in accordance to the item b) of the question statement

(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).

The time of use that is exceeded with probability 0.9 is Enter your answer in accordance to the item c) of the question statement

Answer 1

Solution :

Let X be a random variable which represents the battery life of laptop.

Mean (μ) = 8.5 hours

SD (σ) = 39 minutes = 39/60 = 0.65 hours

It is known that, X ~ N(8.5, 0.65²)

a) We have to find P(X ≥ 9.0 hours).

We know that if X ~ N(μ, σ²) then,

Using "pnorm" function of R we get, P(Z ≥ 0.7692) = 0.2209

Hence, the required probability is 0.2209.

b) We have to find P(X < 8.0 hours).

We know that if X ~ N(μ, σ²) then,

Using "pnorm" function of R we get, P(Z < -0.7692) = 0.2209

Hence, the required probability is 0.2209.

c) Let the time of use that is exceeded with probability 0.9 is k.

Hence, P(X > k) = 0.9

We know that if X ~ N(μ, σ²) then,

......................(1)

Using "qnorm" function of R we get, P(Z > -1.2816) = 0.9

Comparing, P(Z > -1.2816) = 0.9 and (1) we get,

The time of use that is exceeded with probability 0.9 is 7.67 hours.