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.

Answers: 2

Show answers

Answers

answered: 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

answered: Guest

hope it

step-by-step explanation:

x^(1/6) · x^(1/6) · x^(1/6) ·x^(1/6) = x^(4/6) since we add the exponents

4/6 can be simplified to 2/3

the simplified form is x^(2/3)

answered: Guest

step-by-step explanation:

i think something is missing in your question because y-1=(x-12) would not pass through the point (-4,-3). because if you replace x by -4 your are supposed to find y=-3 which is not the case her. if you replace x by -4 you find =-15.

i assume the right equation is y-1=1/4*(x-12) because this one is passing through both your points (if you are interested to know how i found that out i can tell you).

in this equation if i replace x by -4 i find y=-3 and if i replace x by 12, i find y=1 so the line representing this equation is passing through those 2 points.

then the standard form of the equation is finding an expression of y in function of x

y-1=1/4*(x-12)

y=x/4-3+1

y=x/4-2

hope that .

do you confirm there was a mistake in the question?