Question

Suppose that for a specic model of cell phone, the life-length of a fully-charged cellphone battery is normally distributed with a mean of 1200 minutes and a standard deviation of 160 minutes.

(a) A person charges his/her cellphone at night and uses it during the day for about 16 hours. What is the probability that he/she does not have to charge his/her cell phone during the day?

(b) 80% of fully-charged cell-phone batteries have a life-length more than how many minutes?

(c) 30% of fully-charged cell-phone batteries have a life-length between 1250 minutes and how many minutes?

Answer 1

(a) Required probability = P(X > 16*60 minutes) = P(X > 960)

= P( ( X - mean) / SD > (960 - mean) / SD)

= P( Z > (960 - 1200) / 160 )

= P( Z > -1.5)

Using standard normal tables, we get

Probability that he/she does not have to charge his/her cell phone during the day= 0.9332

(b) P(X > P₂₀) = 0.80

=> P(X<P₂₀) = 0.20

=>P( ( X - mean) / SD < (P₂₀ - mean) / SD) = 0.20

= P( Z < (P₂₀ - 1200) / 160 ) = 0.20

Using standard normal tables, we get

(P₂₀ - 1200) / 160 = - 0.84162

P₂₀ = 1065.341 minutes

So, 80% of fully-charged cell-phone batteries have a life-length more than 1065.341 minutes.

(c) We have P(X<1250) =

= P( ( X - mean) / SD < (1250 - mean) / SD)

= P( Z < (1250 - 1200) / 160 )

= P( Z < 0.3125)

Using standard normal tables, we get

= 0.62267

P(X < x ) = 0.62267 - 0.30

=> P( ( X - mean) / SD < (x - mean) / SD) = 0.32267

=> P( Z < (x - 1200) / 160 ) = 0.32267

Using standard normal tables, we get

(x - 1200) / 160 = - 0.46025

x = 1126.36 minutes

So, 30% of fully-charged cell-phone batteries have a life-length between 1250 minutes and 1126.36 minutes.