143k views
4 votes
A simple random sample of 85 analog circuits is obtained at random from an ongoing production process in which 21% 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 P ( 14 ≤ X ≤ 20 ) , the probability that between 14 and 20 circuits in the sample are defective. Report your answer to two decimal places of precision.

1 Answer

1 vote

Answer:

63.81% probability that between 14 and 20 circuits in the sample are defective.

Explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:


E(X) = np

The standard deviation of the binomial distribution is:


√(V(X)) = √(np(1-p))

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
\mu and standard deviation
\sigma, the zscore of a measure X is given by:


Z = (X - \mu)/(\sigma)

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that
\mu = E(X),
\sigma = √(V(X)).

In this problem, we have that:


n = 85, p = 0.21.

So


E(X) = np = 85*0.21 = 17.85


√(V(X)) = √(np(1-p)) = √(85*0.21*0.79) = 3.7552

P ( 14 ≤ X ≤ 20 )

Using continuity correction, this is
P(14 - 0.5 \leq X \leq 20 + 0.5) = P(13.5 \leq X \leq 20.5), which is the pvalue of Z when X = 20.5 subtracted by the pvalue of Z when X = 13.5. So

X = 20.5


Z = (X - \mu)/(\sigma)


Z = (20.5 - 17.85)/(3.7552)


Z = 0.71


Z = 0.71 has a pvalue of 0.7611

X = 13.5


Z = (X - \mu)/(\sigma)


Z = (13.5 - 17.85)/(3.7552)


Z = -1.16


Z = -1.16 has a pvalue of 0.1230

0.7611 - 0.1230 = 0.6381

63.81% probability that between 14 and 20 circuits in the sample are defective.

User Ibram
by
7.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.