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Anestetic [448]

3 years ago

10

In a recent survey in a Statistics class, it was determined that only 74% of the students attend class on Fridays. From past dat

a it was noted that 88% of those who went to class on Fridays pass the course, while only 20% of those who did not go to class on Fridays passed the course.
a.What percentage of students is expected to pass the course?
b.Given that a person passes the course, what is the probability that he/she attended classes on Fridays?

1 answer:

Anna007 [38]3 years ago

7 0

Answer:

a) 83%

b) 0.892

Step-by-step explanation:

percentage that attends class on friday = 74%

percentage that pass because they attend class on friday = 88%

percentage that pass but did not go to school on friday = 20%

a) percentage of students expected to pass the course

= (74% x 88%) +(88% x 20%)

= 0.6512 + 0.176

= 0.8272

= 83%

b) If a person passes the course, what is the probability that he/she attended classes on Fridays

= 74% divided by 83%

= 0.892

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What's the next term? Knowing that the first term, a(1), is 6, the next term is found by multiplying this 6 by (-8)^1, obtaining -48.

The next (third) term is found by multiplying this -48 by (-8)^2, obtaining -48*(+64), or - 3072.

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Step-by-step explanation:

Given that:

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How many centimeters of ribbon that 7 friends will get?

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Answer:

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Step-by-step explanation:

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See attachment for the formatted data

Solving (a): The mean

This is calculated as:

$\bar x = \frac{\sum x}{n}$

So, we have:

$\bar x = \frac{242.87 +212.00 +260.93 +284.08 +194.19 +139.16 +260.76 +436.72 +355.36 +.....+250.61}{20}$

$\bar x = \frac{5203.23}{20}$

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$\sigma = \sqrt{\frac{(242.87 - 260.1615)^2 +(212.00- 260.1615)^2+(260.93- 260.1615)^2+(284.08- 260.1615)^2+.....+(250.61- 260.1615)^2}{20 - 1}}$ $\sigma = \sqrt{\frac{94938.80}{19}}$

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