A simple random sample of 90 analog circuits is obtained at random from an ongoi

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(19X28), the probability that between 19 and 28 circuits in the sample are defective.

Report your answer to two decimal places of precision.

P(19X28)=

Given hint: Since the normal distribution is continuous and the binomial probability distribution uses whole numbers, be sure to use a continuity correction of +0.5 or 0.5 when calculating the probability values. Since different software packages compute zscores differently, this standard normal table link may be useful.

Here is the link to the Z table: http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

Please show ALL work, so I can understand how to do problem, thanks!

Explanation / Answer

n = 90, p = 0.25, q = 1 - p = 0.75

= np = 90 * 0.25 = 22.5 and = (npq) = (90 * 0.25 * 0.75) = 4.11

x1 = 18.5 and x2 = 28.5

z1 = (x1 - )/ = (18.5 - 22.5)/4.11 = -0.9732 and z2 = (x2 - )/ = (28.5 - 22.5)/4.11 = 1.4599

P(18.5 < x < 28.5) = P(-0.9732 < z < 1.4599) = 0.7626

Answer: 0.76

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