Notes Ask Your Teache 1/7 points An instructor who taught two sections of engine

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Notes Ask Your Teache 1/7 points An instructor who taught two sections of engineering statistics last term, the first with 25 students and the second with 30, decided to assign a term project. After all projects had been turned in, the instructor randomly ordered them before grading. Consider the first 15 graded projects a) What is the probability that exactly 10 of these are from the second section? (Round your answer to four decimal places.) (b) What is the probability that at least 10 of these are from the second section? (Round your answer to four decimal places.) c) What is the probability that at least 10 of these are from the same section? (Round your answer to four decimal places.) 134 (d) What are the mean value and standard deviation of the number among these 1S that are from the second section? (Round your mean to the nearest whole number and your standard deviation to three decimal places.) projects projects standard deviation (e) What are the mean value and standard deviation of the number of projects not among these firt 15 that are from the second section? (Round your mean to the nearest whole number and your standard deviation to three decimal places.) mean standard deviation projects projects Need Help? Talk to Titer

Explanation / Answer

total number of students = 25+30 = 55

probability that the student is from second section = 30/55 =

0.545455

n=15

X = number of students out of 15 whose projects are graded are from second section.

X follows binomial distribution

P[X=x] = nCx px (1-p)n-x

a. So, P [ X=10] = 0.13584042385

b. P[X>=10] = P[X=10] + ........ + P [ X=15] = 0.24930037902

c. P[at least 10 of them are from same section] = P[at least 10 of them are from section 1] + P[at least 10 of them are from section2]

probability that a student is from section 1 =

P[at least 10 of them are from section 1]= 0.08225689246

required probability =

0.08225689246+0.13584042385 =

0.218097

d. mean = n*probability = 15*

8.181818

sd = sqrt [ n*prob*(1-prob)]

= sqrt(

3.719008)

=

0.545455

n=15

X = number of students out of 15 whose projects are graded are from second section.

X follows binomial distribution

P[X=x] = nCx px (1-p)n-x

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