Question

The probability that a part produced by a certain​ factory's assembly line will be defective is 0.025.

Suppose a sample of 130 parts is taken. Find the following probabilities by using the normal curve approximation to the binomial distribution.

Use the table of areas under the standard normal curve given below.

The probability that exactly 2 parts will be defective is ____.

​(Round to four decimal places as​ needed.)

The probability that no parts will be defective is _____.

​(Round to four decimal places as​ needed.)

The probability that at least 1 part is defective is ____.

Answer 1

n=

130

p=

0.0250

here mean of distribution=μ=np=		3.25
and standard deviation σ=sqrt(np(1-p))=		1.78
for normal distribution z score =(X-μ)/σx

therefore from normal approximation of binomial distribution and continuity correction:

1)

probability that exactly 2 parts will be defective is:

probability =P(1.5<X<2.5)=P((1.5-3.25)/1.78)<Z<(2.5-3.25)/1.78)=P(-0.98<Z<-0.42)=0.3372-0.1635=0.1737

2)

probability that no parts will be defective is:

probability =P(X<0.5)=(Z<(0.5-3.25)/1.78)=P(Z<-1.54)=0.0618

3)probability that at least 1 part is defective is:

probability =P(X>0.5)=P(Z>(0.5-3.25)/1.78)=P(Z>-1.54)=1-P(Z<-1.54)=1-0.0618=0.9382