Question #259276

A sampling scheme for mechanical components from a production line calls for random samples, each consisting of eight components. Each component is classified as either good or defective. The results of 50 such samples are summa- rized in the table below. Number of Defectives Observed Frequency 30 17 From these data estimate the probability that a single component will be defective, Calculate the probabilities of various numbers of defectives in a sample of eight components, and prepare a table to compare predicted probabili- ties according to the binomial distribution with observed relative frequencies for various numbers of defectives in a sample.


1
Expert's answer
2021-11-01T13:11:18-0400

Let x= number of defectives then X Binomial(8,p)X~Binomial (8,p)

Here p is unknown, we estimate p using method of moment i.e.

p=xf50=171+3250=0.46p= \frac{\sum xf}{50}=\frac{17*1+3*2}{50}=0.46

Hence P(X=x)=(8x)(0.46)x(10.46)8x;x=0,1,2,3,...,8P(X = x) = \begin{pmatrix} 8 \\ x \end{pmatrix}(0.46)^x (1 - 0.46)^{8-x} ; x = 0, 1, 2, 3, ..., 8

P(X=0)=(10.46)8=0.0072P(X=1)=8(0.46)(10.46)7=0.0493P(X=2)=(87/2)(0.46)2(10.46)6=0.1469P(X>2)=1P(X=0)P(X=1)P(X=2)=0.7966P(X = 0) = (1 — 0.46)^8 = 0.0072\\ P(X = 1) = 8 * (0.46) * (1 — 0.46)^7 = 0.0493\\ P(X = 2) = (8 * 7/2) * (0.46)^2 * (1 — 0.46)^6 = 0.1469\\ P(X > 2) = 1 — P(X = 0) — P(X = 1) — P(X = 2) = 0.7966

Expected frequencies :

Eo = 50 * P(X = 0) = 50 * 0.0072 = 0.36

E1 = 50 * P(X = 1) = 50 * 0.0493 = 2.465

E2 = 50 * P(X = 2) = 50 * 0.1469 = 7.345

E>2 = 50 * P(X > 2) = 50 * 0.7966 = 39.83


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