Mathematics, 15.04.2020 03:37 bay62

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.

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Answer from: lindseybug

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

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

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

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

Answer from: Quest

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

$\frac{ab}{bc} =\frac{ad}{ab}$

now, we can cross multiply

and we get

$ab\timesab=ad\times bc$

$ab^2=ad\times bc$

so, this could be step to prove it

Seth is using the figure shown below to prove pythagorean theorem using triangle similarity: in the

Answer from: Quest

I’m pretty sure the * means multiplication. so, if you multiply 5*1, it would equal 5. they are basically are-writing 5x1 as 5*1. so they would be… 5 10152025303540455055