Year Number of Dropouts 1 7 2 8 3 6 4 6 5 6 6 4 7 4 8 1 9 2 10 7 11 6 12 1 13 1 14 1 15 3 16 3 17 5 18 4 19 5 20 7 The table giv

Question

3 years ago

6

Year Number of Dropouts 1 7 2 8 3 6 4 6 5 6 6 4 7 4 8 1 9 2 10 7 11 6 12 1 13 1 14 1 15 3 16 3 17 5 18 4 19 5 20 7 The table giv

es 20 years of data on the number of students in a particular school who dropped out in the ninth grade. A sample is taken to include only the last 3 years of this data. With this information, calculate both the population mean and the sample mean for ninth-grade dropouts per year. The population mean is 4.35. The sample mean is 5.33. The population mean is 4.21. The sample mean is 5.33. The population mean is 5.33. The sample mean is 3.75. The population mean is 5.33. The sample mean is 4.21.

Mathematics

2 answers:

nadya68 [22] · Answer 1 · 2022-02-07T10:31:12+03:00

nadya68 [22]3 years ago

6 0

Population mean = (7+8+6+6+6+4+4+1+2+7+6+1+1+1+3+3+5+4+5+7) / 20 = 87/20 = 4.35 <==

sample mean = (4+5+7) / 3 = 16/3 = 5.33 <==

Alenkinab [10] · Answer 2 · 2022-02-07T11:58:32+03:00

<h2>Answer:</h2>

The population mean is 4.35.

The sample mean is 5.33.

<h2>Step-by-step explanation:</h2>

We are given a table as follows:

Year(x) Number of Dropouts(y)

1 7

2 8

3 6

4 6

5 6

6 4

7 4

8 1

9 2

10 7

11 6

12 1

13 1

14 1

15 3

16 3

17 5

18 4

19 5

20 7

∑ y= 87

The mean of the population is:

$Population\ Mean=\dfrac{\sum y}{20}\\\\\\Population\ Mean=\dfrac{87}{20}\\\\\\Population\ Mean=4.35$

Now, the sample is:

Year(x) Number of Dropouts(y)

18 4

19 5

20 7

Now, sample mean is:

$Sample\ Mean=\dfrac{4+5+7}{3}\\\\\\Sample\ Mean=\dfrac{16}{3}\\\\\\Sample\ Mean=5.33$