The frequency distribution below shows the distribution for

The frequency distribution below shows the distribution for checkout time (in minutes) in a department store between 3:00 and 4:00 PM on a Friday afternoon.

Checkout Time (in minutes)

Frequency

Relative Frequency

1.0 - 1.9

       3

2.0 - 2.9

      12

3.0 - 3.9

       0.20

4.0 - 4.9

       3

5.0 -5.9

Total   

      25

1.Complete the frequency table with frequency and relative frequency. Express the relative frequency to two decimal places.                                                                          

2.

What percentage of the checkout times was at least 3 minutes?

3.

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

4.

Does this distribution have positive skew or negative skew? Why?

Solution

1.

For 3.0-3.9, frequency = 0.20*25 = 5.

Thus, as the frequencies must add up to 25, for 5.0-5.9, frequency = 2.

Consider the table:

2.

P(at least 3) = 0.20+0.12+0.08 = 0.40 or 40% [answer]

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3.

Adding up the first two class at the top, 0.12+0.48 already exceeded 0.50. Thus, the median must lie in 2.0-2.9. [ANSWER, 2.0-2.9]

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4.

It has a positive skew, as there are few data on the greater values of T. It peaked at 2.0-2.9, which is on the left part. [ANSWER, POSITIVE SKEW]