Question

A simple random sample of 105 analog circuits is obtained at random from an ongoing production process in which 30% of all circuits produced are defective.

Let ?X be a binomial random variable corresponding to the number of defective circuits in the sample. Use the normal approximation to the binomial distribution to compute ?(26≤?≤36) , the probability that between 26 and 36 circuits in the sample are defective.

Report your answer to two decimal places of precision.

?(26≤?≤36)=

Answer 1

Here we are given that

n=105

Here success means circuit is to be defective

Probability of success =P=30/100

=3/10=0.3

Probability of failure =q= 1-P=1-(3/10)

7/10=0.7

Now calculating value nP=105*0.3

=31.5 which is greater than 5

Also nq=105*0.7=73.5 it is also greater than 5

Since we got nP and nq are greater than 5 we can do normal approximation

Now Mean =nP=105*0.3

=31.5

Standard deviation =√(nPq) =√(105*0.3*0.7)

=4.695

And by continuity correction the interval 26<=X<=36 then we can use +0.5 and -0.5 to the given interval

Then the interval will became as 25.5<X<36.5

It means the area under the graph ( img1)

Then required probability value

P(25.5<X<36.5) = normal cdf (low,high,mean,standard deviation). [BY CALCULATOR]

P(25.5<X<36.5) =0.71 (answer)