# Suppose that a batch of 100 items contains 6 that are defective and 94 that are non-defective. If X is the number of defective items in a randomly drawn sample of 10 items, find (a)P{X = 0} and (b) P {X > 2}.

Question
Suppose that a batch of 100 items contains 6 that are defective and 94 that are non-defective. If X is the number of defective items in a randomly drawn sample of 10 items, find (a)P{X = 0} and (b) P {X > 2}.

2021-01-20
The event { X=0 } is the event in which 10 non-defective are drawn one after another.
It would go down like this:
The probability of the1st item being non-defective is 94/100,because there are 94 non defective items and 100 items total.
The probability of picking the 2nd non-defective is 93/99, because there are now only 93 non-defective items left, and only a total of 99 items left (we took one out, remember?).
The probability of picking the 3rd non-defective items is 92/98,and so on until all 10 items have been picked.
Each of these are independent events, so we can just multiply them together to get the final result:
$$\displaystyle{\left({\frac{{{94}}}{{{100}}}}\right)}{\left({\frac{{{93}}}{{{99}}}}{\left({\frac{{{92}}}{{{98}}}}\right)}{\left({\frac{{{91}}}{{{97}}}}\right)}{\left({\frac{{{90}}}{{{96}}}}\right)}{\left({\frac{{{89}}}{{{95}}}}\right)}{\left({\frac{{{88}}}{{{94}}}}\right)}{\left({\frac{{{87}}}{{{93}}}}\right)}{\left({\frac{{{86}}}{{{92}}}}\right)}{\left({\frac{{{85}}}{{{91}}}}\right)}={\left({\frac{{{32808912827897606400}}}{{{62815650955529472000}}}}\right)}\approx{0.522}\right.}$$

### Relevant Questions

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
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The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver.
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(d) What point has the property that only 10% of the soil samples have bulk density this high orhigher?
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$$\begin{array}{|c|c|}\hline 11.8 & 7.7 & 6.5 & 6 .8& 9.7 & 6.8 & 7.3 \\ \hline 7.9 & 9.7 & 8.7 & 8.1 & 8.5 & 6.3 & 7.0 \\ \hline 7.3 & 7.4 & 5.3 & 9.0 & 8.1 & 11.3 & 6.3 \\ \hline 7.2 & 7.7 & 7.8 & 11.6 & 10.7 & 7.0 \\ \hline \end{array}$$
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MPa
State which estimator you used.
$$x$$
$$p?$$
$$\frac{s}{x}$$
$$s$$
$$\tilde{\chi}$$
b) Calculate a point estimate of the strength value that separates the weakest $$50\%$$ of all such beams from the strongest $$50\%$$.
MPa
State which estimator you used.
$$s$$
$$x$$
$$p?$$
$$\tilde{\chi}$$
$$\frac{s}{x}$$
c) Calculate a point estimate of the population standard deviation ?. $$[Hint:\ ?x_{i}2 = 1859.53.]$$ (Round your answer to three decimal places.)
MPa
Interpret this point estimate.
This estimate describes the linearity of the data.
This estimate describes the bias of the data.
This estimate describes the spread of the data.
This estimate describes the center of the data.
Which estimator did you use?
$$\tilde{\chi}$$
$$x$$
$$s$$
$$\frac{s}{x}$$
$$p?$$
d) Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds 10 MPa. [Hint: Think of an observation as a "success" if it exceeds 10.] (Round your answer to three decimal places.)
e) Calculate a point estimate of the population coefficient of variation $$\frac{?}{?}$$. (Round your answer to four decimal places.)
State which estimator you used.
$$p?$$
$$\tilde{\chi}$$
$$s$$
$$\frac{s}{x}$$
$$x$$
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Find $$P(A\cap B)$$ and $$P(A\cup B)$$
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$$P>0.250$$
$$P=0.01$$
$$P<0.001$$
$$P<0.002$$

$$\begin{array}{|c|cc|}\hline & \text{Right-Tail Probability} \\ \hline df & 0.250 & 0.100 & 0.050 & 0.025 & 0.010 & 0.005 & 0.001 \\ \hline 1 & 1.32 & 2.71 & 3.84 & 5.02 & 6.63 & 7.88 & 10.83 \\ 2 & 2.77 & 4.61 & 5.99 & 7.38 & 9.21 & 10.60 & 13.82 \\ 3 & 4.11 & 6.25 & 7.81 & 9.35 & 11.34 & 12.84 & 16.27 \\ 4 & 5.39 & 7.78 & 9.49 & 11.14 & 13.28 & 14.86 & 18.47 \\ 5 & 6.63 & 9.24 & 11.07 & 12.83 & 15.09 & 16.75 & 20.52 \\ 6&7.84&10.64&12.59&14.45&16.81&18.55&22.46 \\ 7&9.04&12.02&14.07&16.01&18.48&20.28&24.32\\ 8&10.22&13.36&15.51&17.53&20.09&21.96&26.12 \\ 9&11.39&14.68&16.92&19.02&21.67&23.59&27.88 \\ 10&12.55&15.99&18.31&20.48&23.21&25.19&29.59 \\ 11&13.70&17.28&19.68&21.92&24.72&26.76&31.26 \\ 12&14.85&18.55&21.03&23.34&26.22&28.30&32.91 \\ 13&15.98&19.81&22.36 & 24.74 & 27.69 & 29.82 & 34.53 \\ 14 & 17.12 & 21.06 & 23.68 & 26.12 & 29.14 & 31.32 & 36.12 \\15 & 18.25 & 22.31 & 25.00 & 27.49 & 30.58 & 32.80 & 37.70 \\ 16 & 19.37 & 32.54 & 26.30 & 28.85 & 32.00 & 34.27 & 39.25 \\ 17 & 20.49 & 24.77 & 27.59 & 30.19 & 33.41 & 35.72 & 40.79 \\ 18 & 21.60 & 25.99 & 28.87 & 31.53 & 34.81 & 37.16 & 42.31 \\ 19 & 22.72 & 27.20 & 30.14 & 32.85 & 36.19 & 38.58 & 43.82 \\ 20 & 23.83 & 28.41 & 31.41 & 34.17 & 37.57 & 40.00 & 45.32 \\ \hline \end{array}$$
Dayton Power and Light, Inc., has a power plant on the Miami Riverwhere the river is 800 ft wide. To lay a new cable from the plantto a location in the city 2 mi downstream on the opposite sidecosts $180 per foot across the river and$100 per foot along theland.
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