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IRINA_888 [86]
3 years ago
6

15. What is the conclusion? a. Statistical interpretation agrees with the intuition: the data in Column C is created with the me

an of 30 which is different from 20, and one of the t-tests detected this difference. b. ​​The statistics does not agree with the intuition: the new data is created with different means and yet, the statistics yields inconclusive statement. c. The P-value is very small and we can conclude that the correlation between the data is significant. d. None of these
Mathematics
1 answer:
Masja [62]3 years ago
5 0

Answer:

THREE ... Statistical interpretation agrees with the intuition: the data in Column C is ... of 30 which is different from 20, and one of the t-tests detected this difference. b. ... the intuition: the new data is created with different means and yet, the statistics ...

Step-by-step explanation:

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A food processor packages orange juice in small jars. The weights of the filled jars are approximately normally distributed with
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Answer:

P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)

And we can find this probability using the complement rule and with excel or the normal standard table:

P(z>1.61)=1-P(z

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(10.5,0.3)  

Where \mu=10.5 and \sigma=0.3

We are interested on this probability

P(X>10.983)

And the best way to solve this problem is using the normal standard distribution and the z score given by:

z=\frac{x-\mu}{\sigma}

If we apply this formula to our probability we got this:

P(X>10.983)=P(\frac{X-\mu}{\sigma}>\frac{10.983-\mu}{\sigma})=P(Z>\frac{10.983-10.5}{0.3})=P(z>1.61)

And we can find this probability using the complement rule and with excel or the normal standard table:

P(z>1.61)=1-P(z

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