Question

QC Problem 1 Specifications for a critical part for a DVD player state that the part should weigh between 22 and 26 ounces. The process that produces the parts yields a mean of 24.5 ounces and a standard deviation of 0.85 ounce. The distribution of the weights of the part is normal. Hint: this is an application of the normal distribution. A similar problem appeared in the reliability chapter assignment. a) What percentage of parts will meet the weight specifications? b) If each production month 1,000,000 units of this part are produced at a cost of $50.00 per unit, what is the expected cost of the discarded units? c) The DVD parts have a service life warranty of three years. The average life of the part is six years with a standard deviation of 1.0 year. What percent of the parts will not meet the service life warranty period? The life of the parts is normally distributed. Hint: this is an application of the normal distribution. A similar problem appeared in the reliability chapter assignment. d) Using the information in part C, give an estimate of the replacement cost due to product failure within the warranty period, if 1,000,000 units are sold. Assume that the cost per unit to produce the part is $50.

Answer 1

Let X be a normal random variable which denotes the weight of a part for a DVD player

Mean, $\mu$ = 24.5 ounces

Standard deviation, $\sigma$ = 0.85 ounces

(a) The percentage of parts that will meet specifications

= P(22 < X < 26)

= P{(22 - 24.5)/0.85 < Z < (26 - 24.5)/0.85}

= P(-2.94 < Z < 1.765)

= 0.9596 = 95.96%

(b) The Expected cost of the discarded units

= $50*1,000,000*(1 - 0.9596)

= $2,020,000

For (c) and (d)

Mean, $\mu$ = 6 years

Standard deviation, $\sigma$ = 1 year

Let Y denote the life of tbe part

(c) Percentage of parts that will not meet the service life warranty period = P(Y < 3)

= P{Z < (3 - 6)/1}

= P(Z < -3)

= 0.0013 = 0.13%

(d) The estimated cost due to product failure within the warranty period = $50 * 1,000,000 * 0.0013 = $65000