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pashok25 [27]
3 years ago
5

A local factory uses a manufacturing process in which 70% of the final products meet quality standards and 30% are found to be d

efective. if three of the products are randomly selected, find the probability that all of them are acceptable.
Mathematics
1 answer:
frosja888 [35]3 years ago
8 0
<span>Answer:
   a manufacturing process has a 70% yield, meaning that 70% of the porducts are acceptable and 30% are defective. If three of the products are randomly selected, find the probability that all of them are acceptable. ---
   Ans: 0.7^3 = 0.3430.</span>
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3 2/3 divided by 2 2/3
Goshia [24]

Answer:

1 3/8

Step-by-step explanation:

3 2/3 = 11/3

2 2/3 = 8/3

11/3 / 8/3 = 11/3*3/8 = 11/8 = 1 3/8

5 0
3 years ago
Read 2 more answers
Suppose that a random sample of 10 newborns had an average weight of 7.25 pounds and sample standard deviation of 2 pounds. a. T
frosja888 [35]

Answer:

z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565  

p_v =P(Z  

a) If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.  

b) \chi^2 =\frac{10-1}{1.96} 4 =18.367  

p_v =P(\chi^2 >18.367)=0.0311

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

Step-by-step explanation:

Assuming this info: "Suppose birth weights follow a normal distribution with mean 7.5 pounds and standard deviation 1.4 pounds"

1) Data given and notation  

\bar X=7.25 represent the sample mean  

s=1.2 represent the sample standard deviation

\sigma=1.4 represent the population standard deviation

n=10 sample size  

\mu_o =7.5 represent the value that we want to test  

\alpha=0.05,0.01 represent the significance level for the hypothesis test.  

z would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

2) State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the true mean is less than 7.5, the system of hypothesis would be:  

Null hypothesis:\mu \geq 7.5  

Alternative hypothesis:\mu < 7.5  

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:  

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (1)  

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

3) Calculate the statistic  

We can replace in formula (1) the info given like this:  

z=\frac{7.25-7.5}{\frac{1.4}{\sqrt{10}}}=-0.565  

4)P-value  

Since is a left tailed test the p value would be:  

p_v =P(Z  

5) Conclusion  

Part a

If we compare the p value and the significance level given \alpha=0.05 we see that p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true average is not significantly less than 7.5.  

Part b

A chi-square test is "used to test if the variance of a population is equal to a specified value. This test can be either a two-sided test or a one-sided test. The two-sided version tests against the alternative that the true variance is either less than or greater than the specified value"

n=10 represent the sample size

\alpha=0.01 represent the confidence level  

s^2 =4 represent the sample variance obtained

\sigma^2_0 =1.96 represent the value that we want to test

Null and alternative hypothesis

On this case we want to check if the population variance increase, so the system of hypothesis would be:

Null Hypothesis: \sigma^2 \leq 1.96

Alternative hypothesis: \sigma^2 >1.96

Calculate the statistic  

For this test we can use the following statistic:

\chi^2 =\frac{n-1}{\sigma^2_0} s^2

And this statistic is distributed chi square with n-1 degrees of freedom. We have eveything to replace.

\chi^2 =\frac{10-1}{1.96} 4 =18.367

Calculate the p value

In order to calculate the p value we need to have in count the degrees of freedom , on this case 9. And since is a right tailed test the p value would be given by:

p_v =P(\chi^2 >18.367)=0.0311

In order to find the p value we can use the following code in excel:

"=1-CHISQ.DIST(18.367,9,TRUE)"

Conclusion

If we compare the p value and the significance level provided we see that p_v >\alpha so on this case we have enough evidence in order to FAIL reject the null hypothesis at the significance level provided. And that means that the population variance is not significantly higher than 1.96.

4 0
3 years ago
What's the numerator for the following rational expression? 8/y+3/y=?/y
GuDViN [60]
Ur answer would be 11/y
3 0
3 years ago
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above swimming pool is leaking after 1/2 hour the pool has lead 7/8 of a gallon of water how many gallons of water per hour is a
jekas [21]
Do a proportion. X/7/8=1/1/2. 7/8 x 1= 7/8. Now do 7/8 divided by 1/2. To do this multiply 7/8 by the reciprocal of 1/2 which is 2/1. 7 x 2= 14 and 8 x 1=8. Improper fraction is 14/8. It is 1 6/8 as a mixed number which is simplified to 1 3/4. Answer is 1 3/4
7 0
3 years ago
In college Dak Prescott had a record of 5 to 2, which represented his
kenny6666 [7]

Answer:

y = 3  x = 5

Step-by-step explanation:

If x is the game number, and y is the amount of touchdown passes for that specific game number, then (5,3) represents the idea that the quarter back threw y = 3 touchdown passes for game number x = 5

So the answer is choice D. This is the only ordered pair that comes from the given table. Something like (2,4) isn't true because the quarterback threw 1 touchdown pass (not 4) in game two.

5 0
3 years ago
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