a) If 10 subscribers are chosen at random, probability that 4 or more plan to make real estate purchases in the next two years is 71.56% P(x>=4)? Excel functions: =1-BINOMDIST(3,10,0.4415,1) b) There are 6 subscribers whose household income would be considered an outlier (with the z-score more than 3 or less than -3). They are: Income Z-score 179700 | | | 3.02257 | 200500 | | | 3.619959 | 201700 | | | 3.654424 | 202400 | | | 3.674528 | 205900 | | | 3.77505 | 322500 | | | 7.123872 | Excel functions: Income Mean: =AVERAGE(Data!G2:G411) Standard Deviation: =STDEV(Data!G2:G411) z-score calculation: =(G2-Sheet2!$B$1)/Sheet2!$B$2 or =(G2-74459.51)/34818.21

c) Age of subscribers approximates normal distribution, and according to empirical rule, 68% of subscribers can be expected to be between ages 26 and 34, 95% of subscribers between ages 22 and 38, and 99% of subscribers between ages 18 and 42. Even though ages of subscribers range from 19 to 42, potential advertisers should be aware that data has the expected value 30.11, mode of 32, and the bulk of the subscribers will be between ages of 26 and 34, which determines the potential target market.

Excel Functions:
Variance=(x-m)^2 * f(x)
Standard Deviation=M5^0.5
Expected Value = Weighted Average= age*f(x), where f(x) is % frequency 68.3% subscribers will be within 1 STD of the mean (30 + / -4) | 95% within 2 std (30 + / - 8) | 99% within 3 std (30 + / - 16)Mode=MODE(A2:A411) |

d) The probability of a subscriber having broadband access given that the subscriber has children (conditional probability) is 63.93% P(children)=219/410=0.5341 P(children and access)=140/410=0.3415 P(access given children)=P(children and access)/P(children)=0.3415/0.5341=0.6393