4_BinomialRandomVariable

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Binomial Random Variable Name:_________________________________ Section: ___________ I: Using the Binomial Formula and Binompdf: 1. Seventy-five percent of all U.S. businesses have a Web site. If you randomly select 10 small businesses, what is the probability that exactly four of them have a Web site? Binomial Formula: P(X=4)=C(10,4) ∗ (0.754) ∗ ((1−0.75)10−4) Binompdf: P(x=4) = 0.0162 2. Fifty percent of working adults spend less than 20 minutes commuting to their jobs. If you randomly select 6 working adults, what is the probability that exactly 3 of them spend less than 20 minutes commuting to work? Binomial Formula: p(x=3)=6C3 ×(0.50)³×(1-0.50) ³ ⁶⁻ Binompdf: P(x=3) = 0.3125 3. The results of a recent survey indicate that 71% of Americans believe that they are overweight. If you randomly select 250 Americans, what is the probability that exactly 178 of them believe that they are overweight. Binomial Formula: P(X=178) =C(250,178) ∗ (0.71178) ∗ ((1−0.71)250−178) Binompdf: P(x=178) = 0.0555

4. A survey indicates that 41% of American women consider reading as their favorite leisure time activity. You randomly select four women and ask them if reading is their favorite leisure-time activity. Draw a number line indicating the possible outcomes and circling the outcomes for which you are calculating the probabilities and fnd the probabilities that: a. fewer than 2 of them will respond Yes. 0.4580 b. at least two of them will respond Yes. 0.5420 5. A surgical technique is performed on 12 patients. You are told there is a 70% chance of success. Find the probability that the surgery is successful for a. less than 7 patients. a. 0.11784 b. more than 8 patients. a. 0.49252 c. at least 5 patients. a. 0.99051

d. fewer than 6 patients. a. 0.0386 e. more than 5 but less than 10 patients. a. .7 f. between 4 and 8 patients inclusive. a. 0.5058 6. Suppose you estimate that approximately 8% of students on campus smoke. You decide to do a study about smoking and you randomly recruit 250 students to participate. a. Estimate the mean and st. deviation for this binomial situation. a. Mean = 20 b. St. Dev = 4.27 b. Would you be surprised if your sample contained 26 smokers? Why or why not? It would not be surprising. This number is within one standard deviation from the mean (20 + 4.27 = 24.27), which means it’s within the range of what we would expect for a normal distribution. Would you be surprised if your sample contained 16 smokers? Why or why not? It would not be surprising. This number is also within one standard deviation from the mean (20 - 4.27 = 15.73), which means it’s within the range of what we would expect for a normal distribution. Would you be surprised if your sample contained 33 smokers? Why or why not?

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It would be surprising. This number is more than two standard deviations from the mean (20 + 2*4.27 = 28.54), which means it’s outside the range of what we would typically expect for a normal distribution.

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