NO LINKS OR ELSE YOU'LL BE REPORTED! Only answer if you're very good at Math.No guessing please.

A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu

telo118 [61]

Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step 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:

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

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:

$n = 110, p = 0.2$

So

$\mu = E(X) = np = 110*0.2 = 22$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952$

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

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

$Z = \frac{25 - 22}{4.1952}$

$Z = 0.715$

$Z = 0.715$ has a pvalue of 0.7626.

X = 17

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

$Z = \frac{17 - 22}{4.1952}$

$Z = -1.19$

$Z = -1.19$ has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

4 0

3 years ago

Fatima highlighted columns 3 and 8 in the multiplication table below to find equivalent ratios.

Luba_88 [7]

Answer:

the answer is 40:15

Step-by-step explanation:

7 0

3 years ago

Which expressions are equivalent to -6+49 + (-69)?

m_a_m_a [10]

Answer:

I'm pretty sure it's none of the above

Step-by-step explanation:

Sorry if you get it wrong :/

3 0

2 years ago

How do you write 18 tens in standard form

Liula [17]

1800 because the 18 cant fit in the tens

7 0

3 years ago

(12x + y + z = 26

mel-nik [20]

Option D. D has the matrix of constants [[12], [11], [4]].

Step-by-step explanation:

Step 1:

With the given equations, we can form matrices to represent them.

The coefficients of x, y, and z form a matrix of order 3 ×3, the variables x, y, and z form a matrix of order 1 ×3 and the constants form a matrix of order 1 ×3.

Step 2:

The linear system A is represented as

$\left[\begin{array}{ccc}12&1&1\\1&-11&0\\1&-1&4\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}26\\17\\23\end{array}\right]$ .

Step 3:

The linear system B is represented as

$\left[\begin{array}{ccc}4&1&1\\1&-11&0\\1&-1&12\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}23\\17\\26\end{array}\right]$ .

Step 4:

The linear system C is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}4\\11\\12\end{array}\right]$ .

Step 5:

The linear system D is represented as

$\left[\begin{array}{ccc}1&1&1\\1&-1&0\\1&-1&1\end{array}\right]$ $\left[\begin{array}{ccc}x\\y\\z\end{array}\right]$ $= \left[\begin{array}{ccc}12\\11\\4\end{array}\right]$ .

Step 6:

Of the four options, the linear system D has the matrix of constants [[12], [11], [4]]. So the answer is option D. D.

4 0

2 years ago