Sample MBA Math Exercises - Statistics

Exercise #1: Variance
Unit sales for new product ABC have varied in the first seven months of this year as follows:

Month	Jan	Feb	Mar	Apr	May	Jun	Jul
Unit Sales	428	391	459	161	410	367	466

What is the (population) variance of the data?

Solution Commentary (2:43) Manual Solution The variance is a measure of dispersion from the mean. Specifically, it is the average of the squared difference from the mean. First, we compute the mean: Mean = (428 + 391 + 459 + 161 + 410 + 367 + 466)/7 Mean = 383.143 Now we can compute the variance from the values and mean. Variance = ((428 - 383.143)2 + (391 - 383.143)2 + (459 - 383.143)2 + (161 - 383.143)2 + (410 - 383.143)2 + (367 - 383.143)2 + (466 - 383.143)2)/7 Variance = 9,289 (rounded to the nearest integer) Excel Solution

Exercise #2: Probability
Let X be a discrete random variable. If Pr(X<5) = 2/9, and Pr(X<=5) = 5/18, then what is Pr(X=5)?

Exercise #3: Linear Regression
Consider the following sample data for the relationship between advertising budget and sales for Product A:

Observation	1	2	3	4	5	6	7	8	9	10
Advertising ($)	60,000	70,000	70,000	80,000	80,000	90,000	100,000	100,000	100,000	110,000
Sales ($)	362,000	416,000	417,000	499,000	485,000	536,000	602,000	623,000	616,000	663,000

What is the slope of the "least-squares" best-fit regression line?

Solution Commentary (5:15) Manual Solution The equation for the best-fit regression line through a set of points is given by y = mx + b where m is the slope and b is the y-intercept. The slope m is given by the covariance of x and y divided by the variance of x. The y-intercept b is given by yavg - m * xavg where yavg is the mean of y, m is the slope, and xavg is the mean of x. We take these definitions as given without explanation and focus on computation. We now compute the various intermediate values needed to define the equation of the regression line. Let's compute first the means of x and y, which in our case are the advertising and sales values, respectively. Mean of x = (60,000 + 70,000 + 70,000 + 80,000 + 80,000 + 90,000 + 100,000 + 100,000 + 100,000 + 110,000)/10 Mean of x = 86,000 Mean of y = (362,000 + 416,000 + 417,000 + 499,000 + 485,000 + 536,000 + 602,000 + 623,000 + 616,000 + 663,000)/10 Mean of y = 521,900 Let's now compute the variance of x: Variance of x = ((60,000 - 86,000)2 + (70,000 - 86,000)2 + (70,000 - 86,000)2 + (80,000 - 86,000)2 + (80,000 - 86,000)2 + (90,000 - 86,000)2 + (100,000 - 86,000)2 + (100,000 - 86,000)2 + (100,000 - 86,000)2 + (110,000 - 86,000)2)/10 Variance of x = 244,000,000 Next we compute the covariance of x and y: Covariance of x and y = ((60,000 - 86,000)*(362,000 - 521,900) + (70,000 - 86,000)*(416,000 - 521,900) + (70,000 - 86,000)*(417,000 - 521,900) + (80,000 - 86,000)*(499,000 - 521,900) + (80,000 - 86,000)*(485,000 - 521,900) + (90,000 - 86,000)*(536,000 - 521,900) + (100,000 - 86,000)*(602,000 - 521,900))/10 Covariance of x and y = 1,518,600,000 We can now put together the pieces of the regression line equation: The slope m = 1,518,600,000/244,000,000 = 6.22 The y-intercept b = 521,900 - 6.22 * 86,000 = -13,344 The value requested in this exercise is the slope, which is 6.22 Excel Solution The Excel solution image omits six of the data points for space reasons.

Exercise #4: Normal Distribution
Suppose the daily customer volume at a call center has a normal distribution with mean 4,600 and standard deviation 950. What is the probability that the call center will get fewer than 3,400 calls in a day?

Click here to view a standard normal (z-score) table, if you know how to use it.

Solution Commentary (2:36) Manual Solution "Standard" normal distributions have a mean of zero and a standard deviation of 1. Fortunately, nonstandard normal distributions can be easily converted into standard normal distributions. Solutions obtained for the standard distribution can be applied to the corresponding nonstandard problem. The motivation for this conversion from standard to nonstandard normal distribution is that there is no simple functional form for obtaining solutions to normal distribution problems. Rather, the manual solution process for standard normal distribution problems with the probability specified requires solving the problem using the normal distribution table of values. The table entries consist of probabilities for the interval from the mean to positive z values. We use the first column to identify the row for the z value integer and tenth place. We then use the first row to identify the column for the z value hundredth place. The probability for the z value is at the intersection of the corresponding row and column. We interpolate for z-values with more than two decimal places. We first convert the value of interest from the x scale and units of the original problem to the corresponding z-value, using the formula: z = (x - m)/s, where x is value of interest, s is the standard deviation and m is the mean, all in the units of the original problem z = (3,400 - 4,600)/ 950 z = -1.263 The clearest way to understand the transformations of the exercise into a form that can be answered using the standard normal distribution table of values is to create a set of pictures that illustrate the transformations. (Note that the pictures below show relationship to the mean and do not reflect the exact z-values of the exercise.) P(x < 3,400) = P(z < -1.263) = = 0.5 - P(x < 3,400) = 0.5 - P(0 < z < 1.263) We compute the target probability for a z-value of 1.2632 in a series of steps: 1. Round the target z-value up to the nearest hundredth to find the 'upper' z-value. The 'upper' z-value is 1.27 The corresponding 'upper' probability is 0.3980 2. Note that the 'lower' z-value is 1.26, which is the upper z-value minus 0.01. The corresponding 'lower' probability is 0.3962 3. Because our probability solution is reported to the nearest percent, and the upper and lower probabilities round to the same percent, we can report the probability as 0.40 without further interpolation. P(x < 3,400) = 0.5 - 0.40 = 0.10 Excel Solution