Question

A company manufacturing batteries says that its batteries will last 1400 hours. Suppose that the life times of the batteries are approximately normally distributed, with a mean of 1450 hours and a standard deviation of 60 hours. a) What proportion of the batteries will last less than 1400 hours? b) What is the probability that a randomly selected batteries will last at least 1550 hours? c) How many hours would represent the cutoff for the top 10% of all batteries? d) Would it be unusual for a randomly selected battery to last more than 1600 hours? Why? e) What is the probability that a randomly selected battery will last between 1525 and 1625 hours?

Answer 1

Solution :

Given that ,

mean = = 1450

standard deviation = = 60

P(x < 1400) = P[(x - ) / < ( 1400 - 1450 ) / 60 ]

= P( z < -0.83 )

Using z table,

= 0.2033

Proportion = 0.2033

Solution :

Given that ,

mean = =

standard deviation = =

P(x 1550 ) = 1 - P(x   1550)

= 1 - P[(x - ) / ( 1550 - 1450 ) / 60 ]

= 1 -  P( z 1.67)   

  Using z table,

= 1 - 0.9525

= 0.0475

Probability = 0.0475

The z - distribution of the 10% is

P( Z > z) = 10%

= 1 - P ( Z < z ) = 0.10

= P ( Z < z ) = 1 - 0.10

= P ( Z < z ) = 0.90

P ( Z < 1.282 ) = 0.90

z = 1.282

Using z-score formula,

x = z * +

x = 1.282 * 60 + 1450

x = 1525.92

x = 1526

P(x >1600 ) = 1 - P( x< 1600 )

=1- P[(x - ) / < ( 1600 - 1450) / 60 ]

=1- P( z < 2.5 )

Using z table,

= 1 - 0.9938

= 0.0062

P( 1525 < x < 1625 ) = P[( 1525 - 1450 ) / 60 ) < (x - ) /  < ( 1625 - 1450 ) / 60 ) ]

= P( 1.25 < z < 2.92 )

= P( z < 2.92 ) - P( z < 1.25 )

Using z table,

= 0.9982 - 0.8944

= 0.1038

Probability = 0.1038