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BaLLatris [955]
3 years ago
6

Students in a statistics class are conducting a survey to estimate the mean number of units students at their college are enroll

ed in. The students collect a random sample of 47 students. The mean of the sample is 12.3 units. The sample has a standard deviation of 1.9 units.What is the 95% confidence interval for the average number of units that students in their college are enrolled in? Assume that the distribution of individual student enrollment units at this college is approximately normal.(____,______)Your answer should be rounded to 2 decimal places.
Mathematics
1 answer:
GaryK [48]3 years ago
3 0

Answer:

(11.76, 12.84)

Step-by-step explanation:

Given that we can assume that the distribution of individual student enrollment units at this college is approximately normal.

Sample mean =12.3

sample std dev s = 1.9 units

Sample size = 47

Std error = \frac{1.9}{\sqrt{47} } \\=0.277

Z critical value for 95% = 1.96

Margin of error = 1.96*0.277=0.543

Confidence interval =

(12.3-0.543, 12.3+0.543)\\= (11.757, 12.843)

=(11.76, 12.84)

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2000 soldiers stand in a row. Beginning from the left, each soldier calls out a number, 1, 2, 3, 1, 2, 3, 1, 2, 3, 1, 2, 3, and
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Answer:

The number of soldiers that call out a number 3 is 666 soldiers

Step-by-step explanation:

The parameters given are;

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Soldiers calling from the right = 1, 2, 3,.....

Therefore, since the number soldiers calling out the number 3 are 1 in 3 from the left and 1 in 3 from the right, we split the soldiers into 2 groups of 1000, with one group calling from left and the other group calling from the right;

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Similarly the number of 3s called out from the second group = 333

Hence the total number of soldiers that call out a number 3 = 333 + 333 = 666 soldiers.

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