A random sample of 25 customers was chosen in UMUC MiniMart between 3:00 and 4:0

Question

A random sample of 25 customers was chosen in UMUC MiniMart between 3:00 and 4:00 PM on a Friday afternoon. The frequency distribution below shows the distribution for checkout time (in minutes).

Checkout Time (in minutes)

Frequency

Relative Frequency

1.0 - 1.9

2

2.0 - 2.9

8

3.0 - 3.9

4.0 - 5.9

5

Total

25

Complete the frequency table with frequency and relative frequency.

In what class interval must the median lie? Explain your answer

Assume that the largest observation in this dataset is 5.8. Suppose this observation were incorrectly recorded as 8.5 instead of 5.8. Will the mean increase, decrease, or remain the same? Will the median increase, decrease or remain the same? Why?

Checkout Time (in minutes)

Frequency

Relative Frequency

1.0 - 1.9

2

2.0 - 2.9

8

3.0 - 3.9

4.0 - 5.9

5

Total

25

Explanation / Answer

I begin with the last Q. so if you change the data the mean is the same, because the total datas are equal

and the median remain the same, if you do the same process you always the median is between 3.0-3.9

complete table

The cumulative frequency to the previous class to the middle class is F2 = 10, so that the difference n / 2 - F2 = 2.5, is the number of observations that remain to reach the position of the median.
Md = 3.0 + d
Equating the reasons: d / 2 = 2.5 / 10 = 0.25

so: Md = 3.0 + 0.25 = 3.25

Checkout time Frequency Relative Frequency Acumulate Frequency 1 1.0-1.9 2 0.08 2 2 2.0-2.9 8 0.32 10 3 3.0-3.9 10 0.4 20 4 4.0-5.9 5 0.2 25 Total 25 1 the mean n/2= 12.5 the number 12.5 is the class 3.0-3.9 that class have : Li=3.0 f3=10 C=0.9

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