Find the distance between each pair of points (8,5) ,(-1,3)
2 answers:
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Answer:
5.44% probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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.
Binomial probability distribution:
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
In which is the number of different combinations of x objects from a set of n elements, given by the following formula.
And p is the probability of X happening.
Percentage of parts with weights exceeding 45g?
1 subtracted by the pvalue of Z when X = 45. So
We have
has a pvalue of 0.6915
1 - 0.6915 = 0.3075
If you slect 16 parts at random form that batch, what is the probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g?
This is P(X = 8) when n = 16, p = 0.3075. So
5.44% probability that exactly 8 of the 16 parts you selected will have weights exceeding 45g
The 99% confidence interval for the population proportion is given by
Given that 52% = 0.52 is the standard proportion and the standard error is 2.4% = 0.024.
Therefore, the <span>99% confidence interval for the fraction of u.s. adult twitter users who get some news on twitter is given by
</span>Therefore, the <span>99% confidence interval for the fraction of u.s. adult twitter users who get some news on twitter is 45.82% and 58.18%.
The interpretation of the confidence interval</span> is as follows: "We are 99% confident that the fraction of U.S. adult twiter users who get some news on twitter is between 45.82% and 58.18%.
Given that 52% = 0.52 is the standard proportion and the standard error is 2.4% = 0.024.
Therefore, the <span>99% confidence interval for the fraction of u.s. adult twitter users who get some news on twitter is given by
</span>Therefore, the <span>99% confidence interval for the fraction of u.s. adult twitter users who get some news on twitter is 45.82% and 58.18%.
The interpretation of the confidence interval</span> is as follows: "We are 99% confident that the fraction of U.S. adult twiter users who get some news on twitter is between 45.82% and 58.18%.