Answer:

CLASS MIDPOINT

(xm)

1.94

2.103

3.112

4.121

5.130

f.xm

1.564

2.2,472

3.4,816

4.3,388

5.1,170

cf

1.6/108

2.22/108

3.43/108

4.28/108

5.9/108

Step-by-step explanation:

I HOPE I HELP

Answer:

Solution:

frequency = 6,22,43,29,9

we know that

[tex] mean\frac{sum \: of \: data}{sum \: of \: terms} [/tex]

[tex]mean \frac{6 + 22 + 43 + 29 + 9}{5} [/tex]

[tex] = \frac{6 + 9 + 22 + 28 + 43}{5} [/tex]

[tex] = \frac{108}{5} = 21.6[/tex]

[tex]mean = 21.6[/tex]

[tex] median\frac{(n + 2)}{2} { + }^{4} [/tex]

[tex] = \frac{5 + 1}{2} = \frac{6}{2} = 3[/tex]

so, median is 43

Step-by-step explanation:

Hope it helps

Sorry for incomplete answer

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