Question

1. Your company purchases an expensive component from a supplier. You know that the population of the components is normally distributed with a mean ( μ) of 60 months and a standard deviation (σ) of 2 months. Using this information determine:

1. the probability of a component lasting longer than 68 months.
2. the probability of a component lasting less than 63 months.
3. the probability that a component will last between 58 and 65 months.

2. Using the information in the problem above, what is the probability that a sample average (X) from a sample of 25 components will be between 58.5 and 59.6 months?

Answer 1

Solution :

Given that ,

mean = = 60

standard deviation = = 2

a) P(x > 68) = 1 - P(x < 68)

= 1 - P[(x - ) / < (68 - 60) / 2)

= 1 - P(z < 4)

= 1 - 1

0.0000

Probability = 0.0000

b) P(x < 63) = P[(x - ) / < (63 - 60) /2 ]

= P(z < 1.5)

= 0.9332

Probability = 0.9332

c) P(58.5 < x < 59.6 ) = P[(58.5 - 60)/ 2) < (x - ) /  < (59.6 - 60) / 2) ]

= P(-0.75 < z < -0.2)

= P(z < -0.2) - P(z <-0.75 )

= 0.4207 - 0.2266

0.1941

Probability = 0.1941

d) n = 25

= / n = 2 / 25 = 0.4

= P[(58.5 - 60) /0.4 < ( - ) / < (59.6 - 60) /0.4 )]

= P(-3.75 < Z < -1)

= P(Z < -1) - P(Z < -3.75)

= 0.1587 - 0.0001

0.1586

Probability = 0.1586