Question

The lifetime of batteries in a certain cell phone brand is normally distributed with a mean of 3.25 years and a standard deviation of 0.8 years.
(a) What is the probability that a battery has a lifetime of more than 4 years?
(b) What percentage of batteries have a lifetime between 2.8 years and 3.5 years?
(c) A random sample of 50 batteries is taken. The mean lifetime L of these 50 batteries is recorded. What is the probability that L is less than 3 years?

Answer 1

Solution :

Given that ,

mean = = 3.25

standard deviation = = 0.8

P(x >4 ) = 1 - P(x<4 )

= 1 - P[(x -) / < (4 -3.25) / 0.8]

= 1 - P(z <0.94 )

Using z table

= 1 - 0.8264

= 0.1736

probability= 0.1736

(b)

P(2.8< x <3.5 ) = P[(2.8 -3.25) /0.8 < (x - ) / < (3.5 -3.25) /0.8 )]

= P( -0.56< Z <0.31 )

= P(Z < 0.31) - P(Z < -0.56)

Using z table   

= 0.6217 -0.2877

=0.3340

answer=33.4%

(C)

n = 50

= 3.25

=  / n = 0.8 / 50=0.1131

P( < 3) = P[( - ) / < (3 -3.25) / 0.1131]

= P(z <-2.21 )

Using z table  

= 0.0136

probability=0.0136