Question

An oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 88 hours and a standard deviation equal to 12 hours. (Round your answers to four decimal places.)

(a)

What proportion of the company's drill bits will fail before 79 hours of use?

(b)

What proportion will last at least 79 hours?

(c)

What proportion will have to be replaced after more than 97 hours of use?

Answer 1

Solution :

Given that ,

mean = = 88 hours

standard deviation = = 12 hours

a) P(x < 79) = P[(x - ) / < (79 - 88) / 12]

= P(z < -0.75)

Using z table,

= 0.2266

b) P(x 79) = 1 - P(x   79)

= 1 - P[(x - ) / (79 - 88) / 12 ]

= 1 -  P(z -0.75)   

  Using z table,

= 1 - 0.2266

= 0.7734

c) P(x > 97) = 1 - p( x< 97)

=1- p P[(x - ) / < (97 - 88) / 12 ]

=1- P(z < 0.75)

Using z table,

= 1 - 0.7734

= 0.2266