Qnt 561

Pages: 12 (2214 words) Published: June 19, 2011
Chapter 3; 80, 82, 87

80. Creek Ratz is a very popular restaurant located along the coast of northern Florida. They serve a variety of steak and seafood dinners. During the summer beach season, they do not take reservations or accept “call ahead” seating. Management of the restaurant is concerned with the time a patron must wait before being seated for dinner. Listed below is the wait time, in minutes, for the 25 tables seated last Saturday night.

|28 |39 |23 |67 |37 |28 |56 |40 |28 |50 | |51 |45 |44 |65 |61 |27 |24 |61 |34 |44 | |64 |25 |24 |27 |29 | | | | | |

a. Explain why the times are a population.

These times represent a population because they are representative of all of the tables seated on a particular evening at this restaurant. The definition of a population is an entire collection of items from which one can collect and formulate data. From the collection of times above, one can begin to analyze different aspects such as mean, median and standard deviation.

b. Find the mean of the times.

The mean can be found by adding all of the values of the times and dividing that by the number of times.

- 28+39+23+67+37+28+56+40+28+50+51+45+44+65+61+27+24+61+34+44+64+25+24+27+29

= 1021/25

= 40.84

= 41 minutes was the average wait time

b.2. Find the median of the times.

The median is defined as the midpoint of all of the values after they have been ordered from smallest to largest, or largest to smallest.

- 23,24,24,25,27,27,28,28,28,29,34,37,39,40,44,44,45,50,51,56,61,61,64,65,67

- The median of the wait times is 39 minutes.

c. Find the range of the times.

Range is defined as the difference between the largest and the smallest values in the data set.

QNT 561 – Week 1 Problem Set 3

- The largest value in the set of times is 67 and the smallest is 23.

- 67-23

= 44

d. Find the standard deviation of the times.

To do this, I first took the mean of the times (41 minutes) and subtracted it from each individual midpoint time. The next step was to square each difference and add all of them to get one total sum. Next, I took the total sum of squared differences and divided it by the number of times (25) minus 1. The final step was to take the square root of this total to get the standard deviation.

- 5292 (total of squared differences)

- 5292/25-1 (total of squared differences/number of times – 1)

- 220.5 (take the square root of this number)

= 14.85 minutes is the standard deviation

82. The following frequency distribution reports the electricity cost for a sample of 50 two-bedroom apartments in Albuquerque, New Mexico during the month of May last year.

|Electricity Cost |Frequency |
|\$ 80 up to \$100 |3 |
|100 up to 120 |8 |
|120 up to 140 |12 |
|140 up to 160 |16 |
|160 up to 180 |7 |
|180 up to 200 |4 |
|    Total |50 |

QNT 561 – Week 1 Problem Set 4

a. Estimate the mean cost.

To find the mean, I must first compute the midpoint by adding the upper and lower values in each class of electricity costs.

- (80+100)/2 = \$90

- (100+120)/2 = \$110

- (120+140)/2 = \$130

- (140+160)/2 = \$150

- (160+180)/2 = \$170

- (180+200)/2 = \$190

I then take each midpoint and multiply it by the respective frequency.

- 90*3 = 270

- 110*8 = 880

- 130*12 = 1560

- 150*16 = 2400

- 170*7 = 1190

- 190*4 = 760

Next, I add the total value of all midpoints. I then divide this number by the total frequency count.

- \$7060/50

= \$141.20 is the mean electricity cost.

b. Estimate the standard deviation.

To do this, I first took the mean of the costs (\$141.20) and subtracted it from...