A manufacturer finds that on the average, a television set is used 1.8 hours per day. A one year warranty is offered on the picture tube having a mean time to failure (MTTF) of 2000 hours. If the distribution of time to failure is exponential, then d

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A manufacturer finds that on the average, a television set is used 1.8 hours per day. A one year warranty is offered on the picture tube having a mean time to failure (MTTF) of 2000 hours. If the distribution of time to failure is exponential, then determine the percentage of tubes failing during the warranty period. (15 marks)

The number of defects on 20 items were recorded as given above:

| Item No. | No. of defects | Item No. | No. of defects |
|----------|----------------|----------|----------------|
| 1        | 2              | 11       | 6              |
| 2        | 0              | 12       | 0              |
| 3        | 4              | 13       | 2              |
| 4        | 1              | 14       | 1              |
| 5        | 0              | 15       | 0              |
| 6        | 8              | 16       | 3              |
| 7        | 0              | 17       | 2              |
| 8        | 1              | 18       | 1              |
| 9        | 2              | 19       | 0              |
| 10       | 0              | 20       | 2              |

Use a suitable control chart to identify whether the process is in control or not? (15 marks)

Explain the concepts of producer's and consumer's risks. It has been decided to sample 100 items at random from each large batch. We reject the batch if more than 2 defectives are found. If the acceptable quality level is 1% and the unacceptable quality level is 5%, then find the producer's and consumer's risks. (20 marks)

UPSC Answer Check · Accepted Answer

(a) Let T be the time to failure in hours. T is exponential with MTTF = 2000 hours, so rate λ = 1/2000 h⁻¹.
Warranty usage = 1.8 × 365 = 657 hours.
Using the exponential CDF, P(T ≤ t) = 1 − exp(−λt).
Therefore P(failure during warranty) = 1 − exp(−657/2000) = 1 − exp(−0.3285).
Now exp(−0.3285) = 0.720003. Hence P = 0.279997.
So the percentage of tubes failing during warranty is about 27.9997%, i.e. 28.0%.
Validity: constant average usage of 1.8 h/day and exponential failure law.

(b) Since the number of defects per item is recorded on a constant inspection unit, a c-chart is suitable.
Total defects = 2+0+4+1+0+8+0+1+2+0+6+0+2+1+0+3+2+1+0+2 = 35.
Mean defects per item: c̄ = 35/20 = 1.75.
For a c-chart, UCL = c̄ + 3√c̄ and LCL = c̄ − 3√c̄.
√1.75 = 1.3228757, so 3√c̄ = 3.968627.
UCL = 1.75 + 3.968627 = 5.718627 ≈ 5.72.
LCL = 1.75 − 3.968627 = −2.218627, taken as 0 because defects cannot be negative.
Comparing observations with limits, item 6 has 8 defects and item 11 has 6 defects, both exceeding UCL = 5.72.
Hence the process is not in control.

(c)(i) Producer’s risk α is the probability of rejecting a lot when its quality is actually acceptable, i.e. when p equals the acceptable quality level AQL. Consumer’s risk β is the probability of accepting a lot when its quality is actually unacceptable, i.e. when p equals the unacceptable quality level, also called LTPD.
Here n = 100, reject if more than 2 defectives are found, so accept if X ≤ 2. AQL = 1% = 0.01, LTPD = 5% = 0.05.
Since the batch is large, X is taken as Bin(100, p).

(c)(ii) Producer’s risk at p = 0.01:
α = P(reject | p = 0.01) = P(X > 2) = 1 − P(X ≤ 2).
P(X ≤ 2) = P₀ + P₁ + P₂.
P₀ = (0.99)¹⁰⁰ = 0.3660323413.
P₁ = C(100,1)(0.01)(0.99)⁹⁹ = 0.3697296376.
P₂ = C(100,2)(0.01)²(0.99)⁹⁸ = 0.1848648188.
Thus P(X ≤ 2) = 0.3660323413 + 0.3697296376 + 0.1848648188 = 0.9206267977.
α = 1 − 0.9206267977 = 0.0793732023.
So producer’s risk = 7.94%.

Consumer’s risk at p = 0.05:
β = P(accept | p = 0.05) = P(X ≤ 2).
P₀ = (0.95)¹⁰⁰ = 0.0059205292.
P₁ = C(100,1)(0.05)(0.95)⁹⁹ = 0.0311606801.
P₂ = C(100,2)(0.05)²(0.95)⁹⁸ = 0.0811817719.
Thus β = 0.0059205292 + 0.0311606801 + 0.0811817719 = 0.1182629812.
So consumer’s risk = 11.83%.

Final answers:
(a) 28.0% tubes fail during warranty.
(b) Process is not in control, because items 6 and 11 exceed UCL = 5.72.
(c) Producer’s risk α ≈ 7.94%; consumer’s risk β ≈ 11.83%.

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