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kodGreya [7K]
3 years ago
13

How to solve this problem?

Mathematics
1 answer:
valentinak56 [21]3 years ago
6 0

Very nice handwriting but the math and English are confusing.

Let's assume we're told

\displaystyle 45 = \sum_{i=1}^9 (x_i - 10)^2

The subscript is important.

I think we're told the similar sum with 11 gives the smallest possible value for the sum. This is a rather cagey way of telling us 11 is the mean of the nine points. The mean is the number which minimizes the sum of squared deviations.

\displaystyle 45 = \sum_{i=1}^9 (x_i - 10)^2 = \sum x_i^2 - 20 \sum x_i + 9(100)

\displaystyle  \sum x_i^2=  20 \sum x_i  - 900

If 11 is the mean, the sum of the points is 9(11)=99.

\displaystyle  \sum x_i^2=  20 (99)  - 900 = 1080

Answer: 1080

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Read 2 more answers
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Answer:

0.7995 = 79.95% probability that the sample will contain at least three defectives.

Step-by-step explanation:

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The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

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Normal probability distribution

Problems of normally distributed distributions 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)}.

Suppose that a random sample of 20 items is selected from the machine.

This means that n = 20

The machine produces 20% defectives

This means that p = 0.2

Mean and standard deviation:

\mu = E(X) = np = 20*0.2 = 4

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{20*0.2*0.8} = 1.79

Probability that the sample will contain at least three defectives

Using continuity correction, this is P(X \geq 3 - 0.5) = P(X \geq 2.5), which is 1 subtracted by the pvalue of Z when X = 2.5. So

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

Z = \frac{2.5 - 4}{1.79}

Z = -0.84

Z = -0.84 has a pvalue of 0.2005

1 - 0.2005 = 0.7995

0.7995 = 79.95% probability that the sample will contain at least three defectives.

4 0
3 years ago
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