The mean life of a certain ball bearing can be modeled using a normal distribution with a mean of six years and a standard deviation of one year

12.

Because no probability is given for the switch, we will assume its probability of operating when needed is 100 percent.The expected cost of failure (i.e., without the backup) is$20,000 (1 .98) = $400With the backup, the probability ofnotfailing would be:

.98 + .02(.98) = .9996

Hence, the probability of failure would be1 .9996 = .0004. The expected cost of failure with the backup would be the added cost of the backup component plus the failure cost:

$100 + $20,000(.0004) = $108

Because this is less than the cost without the backup, it appears that adding the backup is definitely cost justifiable.

13.

MTBF = 10 years

Compute the ratioT/MTBF forT= 5, 12, 20, and 30, and obtain the values ofeT/MTBFfromTable 4S.1. The solutions are summarized in the following table:

Question:

The mean life of a certain ball bearing can be modeled using a normal distribution with a mean of six years and a standard deviation of one year. Determine each of the following.a. The probability that a ball bearing will wear out before seven years of service.

b. B. The probability that a ball bearing will wear out after seven years of service (i.e., find its probability)

c. The service life that will provide a wear out probability of 10 percent.

Solution:

Wear-out life men = 6 years

Wear-out standard deviation = 1 year

Wear-out life is normally distributed

a. The probability that a ball bearing will wear-out before seven years of service.

First of all you arte supposed to keep in mind the formula of probability i.e.

z = x- Here

Z = Probability

= Meanx = variable which is T in our case.

To solve the problem (a) we start with given data.

= 6 = 1

T is before seven years

T < 7 _____ (i)

Formula

z = T-

Putting the values

Z = 7-6 1

Z = 1.00

Look the value of probability (z) equal to 1.00 in the table (area under standardized normal curve) which is .8413.(Table is given in tour book Production Operations Manage4ment by William J. Stevenson)See the diagram in book.

b) The probability that a ball bearing will wear out after seven years of service (i.e., find its probability)Now after 7 years we just found before seven years. Total probability in each case is equal to 1 or 100%.

If we subtract the value of before 7 years from total we would get the remaining of after 7 years.

Reliability= Total - probability before 7 years

= 1 -.8413

= .1587

See the diagram on book.

c) The service life that will provide a wear out probability of 10 percent.

In this problem formula is undoubtedly the same.

z = T-

This case is slightly different because we are given the values in %age. We have to find out two variables i.e. T and z.First move towards z.

Look at diagram in book.

Z0.4 = ?

You have to look in the table that what is the probability against 0.4 thats why look in values where a nearer to .4 like .3997 is given and values of z against that is 1.2 on vertical line and 0.08 on horizontal line of probability.

Then methodically you write 1.28 that is value of z.

Put in formula

-1.28 = T - 6

1

T = 6 1.28

= 4.72

See the diagram in book.

The negative sign indicates that value on diagram is on left side or less than mean value which is 6.