Question

in a certain factory turning out razor blades ,there is a small chance 1/1500 for any blade to be defective.The blades are supplied in packets of 10.Use the poisson distribution to calculate the approximate number of packets containing no defective,one defective and two defective blades respectively in a consignment of 10,000 packets given that​​

Answer 1

Answer:

the approximate number of packets $n_3=0$

Step-by-step explanation:

Number of defective blades in a packet has binomial distribution$B(n, p)$ with parameters $n=10$  and $p=0.002$

Binomial distribution can be approximated using Poisson with parameter $m=n \\p=0.02$

Let $X$ equals to number of defective blades in a packet.

$p_{0}=P(X=0)=e^{-0.02}=0.9802$

Using the formula

$p_{x+1}=p_{x} \cdot \frac{m}{x+1}$

we have:

$\begin{gathered}p_{1}=p_{0} \cdot \frac{0.02}{1}=0.019604 \\p_{2}=p_{1} \cdot \frac{0.02}{2}=0.00019604 \\p_{3}=p_{2} \cdot \frac{0.02}{3} \approx 0\end{gathered}$

Thus expected frequencies are:

$\begin{gathered}n_{0}=10000 \cdot p_{0} \approx 9802 \\n_{1}=10000 \cdot p_{1} \approx 196 \\n_{2}=10000 \cdot p_{2} \approx 2 \\n_{3} \approx 0\end{gathered}$

Answer 2

Q- In a factory turning out razor blade, there is a small chance of 1/500 for any blade to be defective. The blades are supplied in a packet of 10. Use Poisson distribution to calculate the approximate number of packets containing blades with no defective, one defective, two defectives and three defectives in a consignment of 10,000 packets.

Given,

The probability of a defective blade = 1 / 500

The blades are supplied in packets of 10

Total packets = 10000

To Find,

The approximate number of packets containing no defective =?

The approximate number of packets containing one defective =?

The approximate number of packets containing two defective =?

Solution,

From binomial distribution, we have B(n,p) where n = 10 and p = 1 / 500

⇒ p = 0.002

The approximation using poisson distribution = m = np = 10 * 0.002

m = 0.02

Le the number of defective blades be x

$P_0= P(X= 0 ) = e^{-0.02}\\P_0= 0.98$

By using formula $P_{x + 1} = P_x(\frac{m}{x + 1} )$, we have

$P_{1} = P_0(\frac{0.02}{0 + 1} )\\P_{1} = 0.0196$

$P_{2} = P_1(\frac{0.02}{1 + 1} )\\P_{2} =0.000196$

$P_{3} = P_2(\frac{0.02}{2 + 1} )\\P_{3} = 0$[Aprrox.]

Hence, approximate number of packets containing no defective,one defective and two defective blades respectively in a consignment of 10,000 packets given that​​ is $P_0$ ≈ 9802 , $P_1$ ≈ 196,  $P_2$ ≈ 2 and $n_3$ ≈ 0.