Business Mathematics: A Complete Guide To Statistics

1. Ungrouped Data
The results of experimentation often yield vast amounts of data. For example, if we wanted to analyze the distribution of income of blue collar workers within a major urban population, we might have several hundred thousand numbers to contend with, each one representing the income of an individual worker. Descriptive statistics is the branch of mathematics which deals with the problem of summarizing data into just a few numbers so that the essence of their behavior can be quickly and easily interpreted.
 

2. The Mean
The mean is a number which measures the “center” of a large collection of numbers called the population. Many people are familiar with the mean as a simple average. It is computed by adding up all the numbers and dividing by their quantity. The Greek letter µ (pronounced “mu”) is used to represent the mean of a population.

where x1 x2...,xN represent the values of the data in the population.


Example 1
To compute the mean of the five numbers 36, 74, 99, 11, and 65 we add them up and divide by 5:


Solved Problems:

11.1   Compute the mean of the numbers 153, 287, 248, 101, 65, and 94.
 

Solution
There are six numbers in the population:


11.2   The annual income of eight families in a small community are $65,000, $74,000, $98,250, $88,500, $57,500, $102,000, $55,350, and $77,000. Compute the mean family income for this community.

The mean family income is $77,200.

11.3   During one trading week the Dow Jones Average closed at 9,139, 9,150, 9,235, 8,911, and 9,050. Compute the mean closing Dow Jones averages for the week.
 

Solution


11.4   Frank filled his Cadillac up with gasoline 5 times during January. The number of gallons he required was 15.2, 13.7, 16.1, 14.9, 16.1.
Compute the mean number of gallons of gasoline his car required.
 

Solution


11.5   A college professor teaches four classes. The number of students in each of her classes is 28, 32, 26, 35.
Her contract with the university states that she must not have a mean class size of more than 30 students. Was this legal under the terms of her contract?
 

Solution
The mean class size for her four classes is

Since this number exceeds 30, her contract was violated.

11.16   A truck driver is entitled to a bonus if the mean number of miles he travels in a week exceeds 100. Rocco traveled the following distances last week:

Is Rocco entitled to his bonus?
 

Solution
The mean number of miles driven is 610/6 = 101.67. Since this exceeds 100 mi, he should get his bonus.

11.7   Don’s delicatessen is open 7 days a week. His gross sales for the first week in February were

What were his mean gross sales for the week?
 

Solution
His mean gross sales for the week were $79,512/7 = $11,358.86.
 

3. The Median
If we were told that the mean of a set of numbers was large, we would probably expect the numbers in the set to be large. The next example shows that this may not be the case at all.


Example 2
Consider the set of numbers 5, 1, 3, 8, 7, 2, 6, 5, 3, 960. The mean of these 10 numbers is

The mean is 100, yet 9 out of 10 of the numbers are less than 10.
Although the mean is an excellent measure of central tendency, it does have one deficiency–its value can be dramatically influenced by the presence of extreme values.
If we were given only the mean, we would be given a misleading representation of the data as a whole . Clearly most of the numbers are small (all except one of them are less than 10), yet the mean has a value of 100.
To allow for a more accurate summary, the median is often given, along with the mean and standard deviation, to describe the data. The median is not affected by the presence of extreme values and can help give a much more realistic picture of the situation.
To compute the median, we first reorder the data, smallest number first and largest last. If the data consist of an odd number of values, the median is the middle value. If the data have an even number of values, we take the two middle values and compute their average. Whether we have an odd or an even number of values, there are always just as many numbers above the median as below it.


Example
Let’s compute the median of the numbers 4, 1, 7, 5, 157. If we reorder (rearrange) the numbers, we get 1, 4, 5, 7, 157.
Since we have an odd number of values, we take the middle value. The median is 5. Note how little an effect the extreme value 157 has on the median.


Example 4
Consider the numbers 7, 4, 4, 2, 6, 100. Reordering, we get 2, 4, 4, 6, 7, 100.
The two middle numbers are 4 and 6. Their average is (4 + 6)/2 = 5. The median is 5.
For a large set of numbers, finding the middle number(s) is easy if we first compute its (their) position(s). Suppose that N represents the size of the data set:
If N is odd, the middle number will have position (N + l)/2.
If N is even, the middle two numbers will have positions N/2 and N/2 + 1.
Always remember to reorder the numbers first!


Example 5
Compute the median of the data sets
(a) 3, 1, 8, 2, 7, 5, 8, 10, 3, 5, 7, 8, 11, 4, 5, 7, 1
(b) 7, 9, 4, 11, 5, 9, 12, 14, 5, 8, 11, 12, 50, 13
Data set (a) has 17 values, an odd number (N = 17). Reordering them, we get 1, 1, 2, 3, 3, 4, 5, 5, 5, 7, 7, 7, 8, 8, 8, 10, 11. The center position is (N + l)/2 = (17 + l)/2 = 9. The median is the 9th value, 5.
Data set (b) has 14 values, an even number (N = 14). Reordering, we get 4, 5, 5, 7, 8, 9, 9, 11, 11, 12, 12,13, 14, 50. The two middle positions are N/2 = 14/2 = 7 and N/2 + 1 = 14/2 + 1 = 8. The corresponding data values are 9 and 11. The median is the average of these two values, which is 10. Note that the median is not necessarily a value of the data set. However, it is the middle number, in the sense that there is an equal number of values above and below it.

Solved Problems

11.8   The heights of six people who work for the same boss are (in inches) 68, 71, 66, 69, 73, and 72. Find their median height.
 

Solution
Reordering the data, we get 66, 68, 69, 71, 72, 73. The two middle values are 69 and 71. Their average is (69 + 71 )/2 = 70. The median height is 70 in.

11.9   The duration of nine telephone calls made from a certain telephone one day was:

Compute the mean and median telephone call duration.
 

Solution
First, we convert the times to seconds. Since there are 60 seconds in a minute, the call times, in seconds, are 67, 131, 97, 191, 125, 210, 126, 146, 50. The mean is

To compute the median, we must reorder the data: 50, 67, 97, 125, 126, 131, 146, 191, 210.
The middle value is the fifth [(9 + l)/2 = 5], so the median call length is 126 s. Note that the median is fairly close to the mean.

11.10  Suppose that an additional telephone call of duration 20 min is added to the calls in Prob. 11.9. What are the new mean and median?
 

Solution
The mean is now

The reordered list of times now reads 50, 67, 97, 125, 126, 131, 146, 191, 210, 1,200. The middle two positions are the fifth (10/2) and sixth (10/2 + 1). Their corresponding values are 126 and 131. The median call length is now (126 + 131)/2 = 128.5. Note how one abnormally large value changes the mean significantly but does not have much effect on the median (compare with Prob. 11.9).

11.11  What would you say about a set of data which has
(a)  A mean of 200 and a median of 20?
(b)  A mean of 20 and a median of 200?
 

Solution
(a)  The data probably have one or more abnormally large values.
(b)  The data probably have one or more abnormally small values.

11.12  Eight bags of sugar are sitting on a shelf. Their weights are

(a)  What is their mean weight?
(b)  What is their median weight?
 

Solution
The weights of the eight bags in ounces are 22, 39, 17, 30, 34, 21, 16, 25. The mean weight of the bags is

To compute the median weight, we must reorder their weights: 16, 17, 21, 22, 25, 30, 34, 39. The median is the average of the two middlemost weights, 22 and 25:


11.13  Compute the median for each of the following data sets:
(a)  41, 54, 42, 53, 49, 51, 37
(b)  41, 54, 42, 53, 49, 51, 37, 75
 
Solution
(a)  We reorder the numbers: 37,41, 42, 49, 51, 53, 54. Since there is an odd number of values N = 7, the middle value is in position (N + l)/2 = 8/2 = 4. The median is the fourth number in the list, 49.
(b)  Now there is an even number of values in the list, N = 8. The middle two positions are N/2 = 4 and N/2 + 1=5, i.e., the fourth and fifth positions. The numbers corresponding to these positions are 49 and 51, so the median is (49 + 51)/2 = 50.
 

4. The Mode
The mode is the value in a set of data that occurs most frequently. There is no mathematical calculation necessary to compute the mode–just look over the data and select the one that appears most often. It often helps to reorder the data first.


Example
To find the mode of the data 15,12,17,15, 21,17,15,27, 9,15,18,17,19,15,11, we first reorder the numbers: 9, 11, 12, 15, 15, 15, 15, 15, 17, 17, 17, 18, 19, 21, 27.
It is clear that the most common number is 15 (repeated 5 times), so the mode = 15.

Solved Problems

11.14  A class of 15 took an exam and received the following grades:

What is the median and the mode of this set of data?
 

Solution
We reorder the data: 52, 53, 69, 75, 75, 75, 78, 80, 80, 82, 85, 86, 90, 92, 97. There are 15 (an odd number) scores: (15 + l)/2 = 8. The eighth score, 80, is the median. The most frequently occurring value is 75. The mode is 75.

11.15  Determine the median and mode of a payroll consisting of 10 salary checks:
$2,150 $1,975 $2,050 $1,900 $2,150
$1,875 $1,750 $1,925 $2,100 $2,175
 

Solution
Reordering the data gives
$1,750 $1,875 $1,900 $1,925 $1,975 $2,050 $2,100 $2,150 $2,150 $2,175
There are 10 (an even number) payroll amounts: 10/2 = 5 and 10/2 + 1 = 6. The average of the fifth and sixth values is (1,975 + 2,050)/2 = 4,025/2 = 2,012.50. The median is $2,012.50. The most common payroll figure is $2,150 (repeated twice). The mode is $2,150.

11.16  Find the median and mode for the following times that it takes a commuter to get from home to the office (all times are in minutes): 24, 36, 22, 24, 28, 30, 32, 24, 22, 26.
 

Solution
Reordering the data is crucial for computing the median and is a convenience for computing the mode: 22, 22, 24, 24, 24, 26, 28, 30, 32, 36. Here, N = 10. Since N is even, we determine the two middlemost values: positions N/2 = 5 and N/2 + 1 = 6. The numbers in the fifth and sixth positions are 24 and 26. The median is the average of these two numbers: (24 + 26)/2 = 50/2 = 25 minutes. The most common value is 24 (repeated 3 times). The mode is 24 minutes.

5. The Standard Deviation
The mean gives an excellent indication of the “center” of a collection of numbers, but gives no information as to how far apart they are spread. Both are necessary if we are to have an accurate summary of the data they represent.

If µ is the mean of the set of numbers x1, x2, ..., xN, then xi – µ1 is called the deviation from the mean. It tells us how far the number xt is from the mean, and on which side of the mean xi lies. The deviation xi – µ is positive if xi is greater than µ and negative if xi is less than µ

The number (xi – µ)2 is called the squared deviation from the mean. It can never be negative.

The average of the squared deviations is called the variance of the data, and is represented by V.

The purpose of squaring is to eliminate negative deviations. If we did not square, the positive and negative deviations would always add to 0. Unfortunately, squaring also exaggerates the values of the deviations, making the variance an unrealistic measure. We compensate by taking the square root of the variance.

This number is called the standard deviation and is represented by the Greek letter a (sigma). (You can easily calculate square roots using the √ key on your calculator.)



Example
Compute the standard deviation of the sets of numbers (a) 3, 5, 7, 9, 11; (b) 1, 3, 7, 10, 14; and (c) 7, 7, 7, 7, 7.

(a) The mean must first be computed:

Next the variance:

Finally the standard deviation:

Even though the mean in parts (a) and (b) are both 7, the standard deviation in (b) is much larger than in (a). This is because the numbers in (b) are farther from the mean, so that the individual squared deviations are larger.

Note once again that the mean is 7. The standard deviation is 0 because all the numbers are equal and identical to the mean; 0 is the smallest value that the standard deviation can ever be. This can happen only when all the numbers are the same.

There is another way to compute the variance, which is a little more convenient, especially when the mean is not a “round” number. The following alternate formula can be shown to be mathematically equivalent to the formula for variance given previously:



Example
To show that this formula is an alternate way of computing the variance, we apply it to the data set of Example 7, part (b). The numbers are 1, 3, 7, 10, 14. We have already seen that µ = 7.

It is sometimes convenient to compute the standard deviation using a tabular format. The next example illustrates how this is done.


Example
To compute the mean and standard deviation for the data in Example 7, part (b) using a tabular format, we form a table with two columns. The first column represents the data and the second column the square of the data.

Then we add the two columns:


Solved Problems

11.17  Compute the mean, variance, and standard deviation of the data 2, 5, 8, 10, 12, 14, 19.
 

Solution
There are seven numbers:


11.18  Use the alternate formula illustrated in Example 8 to compute the variance and standard deviation of the data in Prob. 11.17.
 

Solution


11.19  Compute the mean and standard deviation of the data in Prob. 11.17 using a tabular format.
 

Solution


11.20  The annual salaries (in thousands of dollars) for a group of college professors from eight universities are 65, 57, 60, 75, 76, 72, 64, and 69. Compute their mean income and standard deviation.
 

Solution

Their mean income is $67,250. The standard deviation is $6,476.

11.21  What is the mean and standard deviation of the set of data 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13.
 

Solution
Since the data values are identical, their mean is 13 and the standard deviation is 0.

11.22  A donut machine produces the following quantities of donuts during a 5-hour test. Compute the mean and standard deviation of the numbers of donuts made.

 

Solution


11.23  Without performing any calculation, which data set has the larger (a) mean and (b) standard deviation?

Solution
(a)  50 is the mean of data set 1. (Observe that each number to the left of 50 is “counterbalanced” by a number equally as far to the right of 50.) Similarly, data set 2 has a mean of 50. Both sets have the same mean.
(b)  Since the numbers in data set 2 are farther from the mean than the numbers in data set 1, data set 2 has a larger standard deviation.

11.24  Confirm your suspicions in Prob. 11.23 by doing the actual calculations.

Observe that the means of both data sets are the same. The standard deviation of data set 2 is precisely double that of data set 1. (Can you see why?)

6. Grouped Data
Computing the mean and standard deviation of large groups of data can be difficult. Often, data sets have repetitive values and the formulas for µ and σ can be modified to take advantage of this.

Suppose that the value x is repeated 3 times. Instead of adding x + x + x it is simpler to multiply x by 3 to get 3x.

More generally, if x is repeated f times (f is called the frequency of x), the value equivalent to fx or xf
The mean of a set of numbers can be written

Here, n represents the number of different values in the set and N represents the total number of values. Note that f1 + f2 + ...,+fn. N is called the total frequency.
The mean can be conveniently computed in a tabular format.

Example
Compute the mean of the set of numbers 2, 2, 3, 3, 3, 3, 4, 4, 4, 5. First we list the values of x and their corresponding frequencies:

The variance and standard deviation can be computed in similar fashion. Since it is easily adapted to tabular calculation, the most convenient formula to use is



Example
Compute the variance and standard deviation for the data of Example 10. Recall that µ = 3.3.

When a data set consists of a large number of values, it is convenient to group them into classes. The upper and lower endpoints of each class are called the class limits. The number of values that fall within each class is called the class frequency. The midpoint of each class, called the class mark, may be used together with the class frequency to obtain accurate approximations of the mean and standard deviation.


Example
A sample of 50 students showed the following grades on a college entrance exam:

First we group the data into classes of width 10 starting with a lower class limit of 40. The class mark is the average of the lower and upper class limits.

The frequency is the number of values which fall in each class.


Solved Problems

11.25  Find the mean and standard deviation of the following data:

 

Solution


11.26  One hundred tires were driven until their tread wore out. The number of miles driven, to the nearest 5,000, was determined and tabulated:

Compute the mean and the standard deviation for the number of miles driven on these tires.
 

Solution



11.27  A survey of the ages women marry yielded the following data:

(a)  Starting with a lower class limit of 18, group the data into five classes of width 3.
(b)  Compute the mean.
(c)  Compute the variance and the standard deviation.
 

Solution


 

More Statistics Problems:

11.28  Compute the mean of the seven numbers 123, 235, 342, 411, 252, 373, 399.

11.29  The annual gross income for the Jones family is

Compute the mean annual gross income for the Jones family.

11.30  Compute the mean high temperature for the week given the following daily high temperatures (in degrees Fahrenheit): 70, 72, 62, 79, 69, 70, 75.

11.31  To compute the gas mileage on his new car, Jeremy filled it up 5 times with a measured gallon of gasoline and drove it each time until he ran out of gas. The numbers of miles he traveled were 16.3, 16.9, 16.8, 16.8, and 16.4. What was the mean number of miles Jeremy’s car traveled on a gallon of gas?

11.32  If a salesman travels 157 mi on Monday, 143 mi on Tuesday, 211 mi on Wednesday, 120 mi on Thursday, and 146 mi on Friday, what is his mean distance traveled for the week?

11.33  Mrs. Martin had sextuplets (6 babies). Their weights were 3 lb 6 oz, 3 lb 1 oz, 3 lb 8 oz, 3 lb 7 oz, 2 lb 10 oz, and 3 lb 2 oz. What is the mean weight of Mrs. Martin’s babies?

11.34  The heights of 5 men are 6 ft 0 in, 6 ft 1 in, 5 ft 10 in, 6 ft 0 in, and 5 ft 11 in. Compute the mean, median, and standard deviation of their heights.

11.35  The men in Prob. 11.34 married, respectively, women whose heights are 5 ft 6 in, 5 ft 9 in, 5 ft 7 in, 5 ft 10 in, and 5 ft 11 in. Compute the mean, median, and standard deviation of their heights.

11.36  Compute the mean, median, and standard deviation of the difference in heights between the spouses of problems 11.34 and 11.35.

11.37  What can be said about a set of data whose mean is 149 and whose standard deviation is 0?

11.38  The following three sets of numbers each have mean 60. Which has the largest standard deviation? Which has the smallest standard deviation? (Do not perform any calculation.)


11.39  Compute the standard deviation for each of the data sets of Prob. 11.38 to confirm your previous answer.

11.40  What would happen to the mean of a set of data if each value were increased by 5?

11.41  What would happen to the variance and standard deviation of a set of data if each value were increased by 5?

11.42  A count of the number of cashew nuts in 10 cans of mixed nuts yielded the following data:
Number of cashews: 27, 30, 28, 24, 22, 28, 30, 28, 25, 29
Compute the mean, median, mode, and standard deviation of the number of cashews found.

11.43  The number of innings in 10 Major League baseball games on a recent Yankee road trip was 9,10, 9, ll, 8, 9, 10, 10, 9, 10. Compute the mean, median, and mode.

11.44  The number of 911 calls arriving at a central dispatch switchboard during 5 consecutive hours was 37, 23, 28, 29, and 32. Compute the mean, median, and standard deviation.

11.45  A class of 20 students took an examination in business mathematics and received the following scores:

Compute the mean, median, and mode.

11.46  Compute the mean, variance, and standard deviation of a set of numbers with the following frequency distribution:


11.47 Twenty-five women were asked to state the age at which they had their first child. The results were put into a frequency table:

Compute the mean age at which each woman’s first child was born. Compute the standard deviation.

11.48  In a medical study, the number of days it took to recover from a certain strain of flu was recorded for a group of 50 people. The results were tabulated and put into a frequency table:

Compute the mean recovery time. What is the standard deviation?

11.49  A sample of 50 oranges taken from an orange tree was sampled to determine their size.

Compute the mean, variance, and standard deviation of the sample.

11.50  Professor Lerner will curve his examination if the mean grade of the class is less than 60. The distribution of grades on his most recent exam was

Will Professor Lerner curve his examination? Use the class marks (center of each class) to approximate the mean.

11.51 A survey was taken of the number of pounds 50 people lost while participating in a nationally advertised weight-loss plan.

Pounds Lost

(a)  Organize the data into a frequency distribution using classes 1-5, 6-10, and so forth.
(b)  Using the class mark (midpoint of each class), compute the mean, variance, and standard deviation.

11.52  The annual incomes of 40 people randomly selected from the workforce were compiled.

Annual Income (in Thousands of $)

(a)  Organize the data into a frequency distribution using classes 20-29, 30-39, and so on.
(b)  Using the class mark (midpoint of each class), compute the mean, variance, and standard deviation.

Answers to Above Statistics Problems:

11.28   305
11.29   $51,104
11.30   71°
11.31   16.64 mi
11.32   155.4 mi
11.33   3 lb 3 oz (51 oz)
11.34   Mean = 5 ft 11.6 in (71.6 in), median = 6 ft 0 in (72 in), standard deviation = 1.02 in
11.35   Mean = 5 ft 8.6 in (68.6 in), median = 5 ft 9 in (69 in), standard deviation = 1.85 in
11.36   Mean = 3 in, median = 3 in, standard deviation = 2 in
11.37   All numbers in the data set are 149
11.38   Set C has largest standard deviation; set A has smallest
11.39   Set A, σ = 5.66; set B, σ = 7.07; set C, σ = 14.14
11.40   The mean would increase by 5
11.41   No change in standard deviation since mean increases along with each value of the data; deviation from mean remains unchanged
11.42   µ = 27.1, median = 28, mode = 28, σ = 2.51
11.43   µ = 9.65, median = 9.5, mode = 9
11.44   µ 29.8, median = 29, σ = 4.62
11.45   µ = 72.4, median = 75, mode = 75
11.46   µ = 4.6, V = 2.54, σ = 1.59
11.47   µ = 22.4 years, σ = 2.24 years
11.48   µ = 3.22 days, σ = 1.14 days
11.49   µ = 8.964 cm, V = 0.01, σ = 0.10 cm
11.50   Class marks = 44.5, 54.5, 64.5, 74.5, 84.5, 94.5 using these values, µ = 68.75, so Professor Lerner will not curve the exam
11.51   (a)

(b) µ = 10 lb, V = 45, σ = 6.71 lb

11.52   (a)

(b) µ = 56.25 ($56,250), V = 359.44, σ = 18.959 ($18,959).

Note: The use of xi is standard mathematical notation. i is a “dummy” variable which can be any number from 1 to N, so xi represents any of the numbers x1 x2, ...,xN. Using this device, we can discuss the data more conveniently.