Question

One-fifth per cent 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 the approximate number of packets containing no defective in a consignment of1,00,000 packets.

Answer 1

no defective blades in a consignment of 1,00,000 packets is 99837.

Explanation:

we first need to find the mean number of packets containing no defective blades. This mean is equal to the total number of packets (1,00,000) multiplied by the probability of a packet containing no defective blades.

The probability of a blade being defective is given as 0.002 (or 2/1000) or 2/100 or 1/50 or 2%.

Since a packet contains 10 blades. So the probability of a packet containing no defective blades is (1 - 0.002)^10 = 0.999936 or (1-0.02)^10 = 0.99837

So the mean number of packets containing no defective blades is (1,00,000 * 0.99837) = 99837

Answer 2

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