5-2 Binomial Probability Distributions 225 a. With n = 460, p = 0.5, and q = 0.5, Formulas 5-6 and 5-8 can be applied as follows: m = np = 1460210.52 = 230.0 games s = 1npq = 21460210.5210.52 = 10.7games 1rounded2 For random groups of 460 overtime games, the mean number of wins is 230.0 games, and the standard deviation is 10.7 games. b. The values separating numbers of wins that are significantly low or significantly high are the values that are two standard deviations away from the mean. With m = 230.0 games and s = 10.7 games, we get Significantly low … m - 2s = 230.0 - 2110.72 = 208.6 games Significantly high Ú m + 2s = 230.0 + 2110.72 = 251.4 games Significantly low numbers of wins are 208.6 games or fewer, significantly high numbers of wins are 251.4 games or greater, and values not significant are between 208.6 games and 251.4 games. c. The result of 252 wins is significantly high because it is greater than the value of 251.4 games found in part (b). SOLUTION YOUR TURN. Do Exercise 29 “Gender Selection.” Instead of the range rule of thumb, we could also use probabilities to determine when values are significantly high or low. Using Probabilities to Determine When Results Are Significantly High or Low ■ Significantly high number of successes: x successes among n trials is significantly high if the probability of x or more successes is 0.05 or less. That is, x is a significantly high number of successes if P(x or more) … 0.05.* ■ Significantly low number of successes: x successes among n trials is significantly low if the probability of x or fewer successes is 0.05 or less. That is, x is a significantly low number of successes if P(x or fewer) … 0.05.* *The value 0.05 is not absolutely rigid. Other values, such as 0.01, could be used to distinguish between results that are significant and those that are not significant. Go Figure $5: Cost of the ticket on the first commercial airline flight in the United States in 1914, which flew 21 miles. Rationale for the Binomial Probability Formula The binomial probability formula is the basis for all three methods presented in this section. Instead of accepting and using that formula blindly, let’s see why it works. In Example 2, we used the binomial probability formula to find the probability of getting exactly two adult smartphone owners who are cashless (never use cash) when ten adult smartphone owners are randomly selected. With P1one person is cashless2 = 0.05, we can use the multiplication rule from Section 4-2 to find the probability that the first two adults are cashless and the last eight adults are not cashless. We get the following result: P12 adults are cashless followed by 8 adults who are not cashless2 = 0.05# 0.05# 0.95# 0.95# 0.95# 0.95# 0.95# 0.95# 0.95# 0.95 = 0.052 # 0.958 = 0.00165855 This result gives a probability of randomly selecting ten adult smartphone owners and finding that the first two are cashless and the last eight are not. However, the probability of
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