One fifth percent of the blades produced by a blade manufacturing factory turn out to be defective. The blades are supplied in packets of 10. Use Poisson distribution to calculate approximate number of packets containing no defective, one defective respectively in a consignment of 1,00,000 packets.
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Probability of defect per blade = 1/500 = 0.002
Poisson distribution :
Pₓ (k) = (x ^k)/k × e⁻ˣ
Where k = The number of defective blades in a packet.
For a packet of 10 blades the mean number of defect x = 0.002 × 10 = 0.02
1.) When k = 0
Pₓ(0) = (0.02⁰ / 0!) × e - 0.02 = 0.980199
The approximate number of packets containing blades with no defective is :
10000 × 0.980199 = 9802
2.) When k = 1
Pₓ(1) = (0.02 / 1!) × e-0.02 = 0.019604
Approximate number containing one defective is :
10000 × 0.019604 = 196
3.) When k = 2
Pₓ(2) = (0.02² / 2!) × e - 0.02 = 0.000196
Approximate number containing 2 defective :
0.000196 × 10000 = 1.9 = 2
4.) When k = 3
Pₓ(3) = (0.02³ / 3!) × e-0.02 = 0.000013
Approximate number containing 3 defective is :
0.000013 × 10000 = 0.13 = 0
Poisson distribution :
Pₓ (k) = (x ^k)/k × e⁻ˣ
Where k = The number of defective blades in a packet.
For a packet of 10 blades the mean number of defect x = 0.002 × 10 = 0.02
1.) When k = 0
Pₓ(0) = (0.02⁰ / 0!) × e - 0.02 = 0.980199
The approximate number of packets containing blades with no defective is :
10000 × 0.980199 = 9802
2.) When k = 1
Pₓ(1) = (0.02 / 1!) × e-0.02 = 0.019604
Approximate number containing one defective is :
10000 × 0.019604 = 196
3.) When k = 2
Pₓ(2) = (0.02² / 2!) × e - 0.02 = 0.000196
Approximate number containing 2 defective :
0.000196 × 10000 = 1.9 = 2
4.) When k = 3
Pₓ(3) = (0.02³ / 3!) × e-0.02 = 0.000013
Approximate number containing 3 defective is :
0.000013 × 10000 = 0.13 = 0
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Answer:
0.02
0.98
0.019
0.00019
20 packets
Explanation:
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