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Natali [406]
3 years ago
9

Suppose that there is a coffee shop in an office(Which also has interns). It is observed that out of the people that visit the c

offee shop in a week 25% were female members, 40% were male members, 25% were female interns and 40% were male interns. These people are randomly asked about the coffee taste.
What is the probability that the first 6 people who are asked about the taste are all office members and not interns?
And what is the probability that the first intern that we talk to is a female?
Mathematics
1 answer:
ser-zykov [4K]3 years ago
8 0
The probability that the first 6 people who are asked that aren't interns is 35% 
The probability that the first intern that we talk to is a female is 25%
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Answer:

The degrees of freedom is 11.

The proportion in a t-distribution less than -1.4 is 0.095.

Step-by-step explanation:

The complete question is:

Use a t-distribution to answer this question. Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution less than -1.4  if the samples have sizes 1 = 12 and n 2 = 12 . Enter the exact answer for the degrees of freedom and round your answer for the area to three decimal places. degrees of freedom = Enter your answer; degrees of freedom proportion = Enter your answer; proportion

Solution:

The information provided is:

n_{1}=n_{2}=12\\t-stat=-1.4

Compute the degrees of freedom as follows:

\text{df}=\text{Min}.(n_{1}-1,\ n_{2}-1)

   =\text{Min}.(12-1,\ 12-1)\\\\=\text{Min}.(11,\ 11)\\\\=11

Thus, the degrees of freedom is 11.

Compute the proportion in a t-distribution less than -1.4 as follows:

P(t_{df}

                      =P(t_{11}>1.4)\\\\=0.095

*Use a <em>t</em>-table.

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