Question

A simple random sample of 90 analog circuits is obtained at random from an ongoing production process in which 25% 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(19 ≤ X ≤ 28), the probability that between 19 and 28 circuits in the sample are defective. Report your answer to two decimal places of precision.

Answer 1

Answer:

76.10% probability that between 19 and 28 circuits in the sample are defective

Step-by-step explanation:

I am going to use the binomial approximation to the normal to solve this question.

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:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$n = 90, p = 0.25$

So

$\mu = E(X) = np = 90*0.25 = 22.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{90*0.25*0.75} = 4.11$

P(19 ≤ X ≤ 28)

Using continuity correction, this is $P(19-0.5 \leq X \leq 28+0.5) = P(18.5 \leq X \leq 28.5)$, which is the pvalue of Z when X = 28.5 subtracted by the pvalue of Z when X = 18.5. So

X = 28.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{28.5 - 22.5}{4.11}$

$Z = 1.46$

$Z = 1.46$ has a pvalue of 0.9279

X = 18.5

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{18.5 - 22.5}{4.11}$

$Z = -0.97$

$Z = -0.97$ has a pvalue of 0.1669

0.9279 - 0.1669 = 0.7610

76.10% probability that between 19 and 28 circuits in the sample are defective