Question

The life of a machine component is normally distributed with a mean of 5,000 hours and a standard deviation of 200 hours. Find the probability that a randomly selected component will last:

1. more than 5,100 hours
2. less than 4,850 hours
3. between 4,950 and 5,200 hours

Answer 1

Solution :

Given that,

mean = = 5000

standard deviation = = 200

a ) P (x > 5100 )

= 1 - P (x < 5100 )

= 1 - P ( x -  / ) < ( 5100 - 5000 / 200 )

= 1 - P ( z < 100 / 200 )

= 1 - P ( z < 0.50 )

Using z table

= 1 - 0.6915

=0.3085

Probability =0.3085

b ) P( x < 4850 )

P ( x - / ) < ( 4850 - 5000 / 200 )

P ( z < -150 / 200 )

P ( z < -0.75 )

= 0.2266

Probability = 0.2266

c ) P (4950 < x < 5200)

P ( 4950 - 5000 / 200 ) < ( x -  / ) < ( 5200 - 5000 / 200 )

P ( - 50 / 200 < z < 200 / 200 )

P (-0.25 < z < 1)

P ( z < 1 ) - P ( z < -0.25)

Using z table

= 0.8413 - 0.4013

= 0.4400

Probability =0.4400