Refer to the following frequency distribution for Questions 2, 3, 4, and 5. Show

Question

Refer to the following frequency distribution for Questions 2, 3, 4, and 5. Show all work. Just the answer, without supporting work, will receive no credit.

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

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

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

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

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

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

Explanation / Answer

2)

3) add the frequencies for 3 and above. so 5+3+5=13.

in terms of percentage (13/25)*100=52%

4) N=25

since median is the middle most value, we find N/2=12.5

look af lcf >=12.5, it is 15. so the class interval 2 to 2.9 is the median class.

we can find the measure of skewness using the formula

s=xbar-Z/sd

xbar=sum(fx)/N

mean=75.25/25=3.01

sd=sqrt(sumfx^2/N-(sumfx/N)^2=sqrt(256.6625/25-(75.25/25)^2=1.0983

Z=L+(fm-f1/2fm-f1-f2)*c

fm is the highest frequecy=12

L=lower of the class with the hghest frequenc=1.95

c=class width=1

z=1.95+(12-3/24-3-5)*1=2.5125

s=3.01-1.0983/2.5125=0.7608

This is a positive value, hence the distirbution is positively skewed.

chekout time frequency relative frequency=class frequency/total frequency 1to 1.9 3 0.12 2 to 2.9 12 0.48 3 to 3.9 25*0.2=5 0.2 4 to 4.9 3 0.12 5 to 5.9 25-(12+5+3+3)=2 0.08 total 25

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