Question

The length of life of an instrument produced by a machine has a normal distribution mean 24 months and standard deviation 3 months. Find the probability that an instrument produced by this machine will last: (a) over 23 months? (b) less than 26 months? (c) between 22 and 25 months?

Answer 1

Given ,

$\mu$ = 24, $\sigma$ = 3

We convert this to standard normal as

P( X < x) = P (Z < x - $\mu$ / $\sigma$ )

a)

P( X > 23) = P( Z > 23 - 24 / 3)

= P (Z > -0.3333)

= P( Z < 0.3333)

= 0.6306

b)

P( X < 26) = P( Z < 26 - 24 / 3)

= P( Z < 0.6667)

= 0.7475

c)

P( 22 < X < 25) = P( X < 25) - P( X < 22)

= P(Z < 25 - 24 / 3) - P( Z < 22 - 24 / 3)

= P( Z < 0.3333) - P( Z < -0.6667)

= P( Z < 0.3333) ( 1 - P( Z < 0.6667) )

= 0.6306 - ( 1 - 0.7475)

= 0.3781