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Jerry has taken a random sample of students and determined the number of electives that each student in his sample took last year. There were 19 students in the sample. Here is the data on the number of electives the 19 students took: 6, 6, 8, 7, 7, 7, 8, 9, 10, 8, 7, 6, 9, 6, 8, 7, 9, 7, 10. The mean of this sample data is 7.63.

What is the sample proportion of students who took fewer than the mean number of electives?
Answers
A.10/19
B.6/19
C.7/19
D.There is not enough data to answer this question.

User Dthree
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2 Answers

6 votes
I would say that the answer would be letter A. 10/19. I hope you are satisfied with my answer and feel free to ask for more!(: 
User Eduard Kolosovskyi
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6 votes

Answer: The correct proportion is (A)
(10)/(19).

Step-by-step explanation: Given that Jerry has taken a random sample of students and determined the number of electives that each student in his sample took last year. There were 19 students in the sample.

The data on the number of electives that 19 students took is

6, 6, 8, 7, 7, 7, 8, 9, 10, 8, 7, 6, 9, 6, 8, 7, 9, 7, 10.

The mean of this sample data is 7.63.

We are to find the number of proportion of students who took fewer than the mean number of electives.

Let 'A' denotes the set of students who took number of electives fewer than the mean number of electives, i.e., 7.63.

So, n(A) = 10.

And, let 'S' denotes set of all students in the sample , then we have

n(S) = 19.

Therefore, the sample proportion of students who took fewer than the mean number of electives is given by


S_p=(n(A))/(n(S))=(10)/(19).

Thus, (A) is the correct option.

User Guypursey
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