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Stat 250.3 October 22, 2003 HOMEWORK 4– SOLUTIONS 8.62 a. Answer = .9015. For a binomial random variable with n = 1000 and p = .60, = np = 1000(.60) = 600, and = 1000 (.60 )(1 .60 ) 15 .492 . 620 600 1.29 . P(Z 1.29) = .9015. 15 .492 b. Answer = .0039. For a binomial random variable with n = 2000 and p = .87, For X = 620, z = np = 2000(.87) = 1740, and = For X = 1700, z 8.84 2000 (.87 )(1 .87 ) 15 .04 . 1700 1740 2.66 . P(Z 2.66) = .0039. 15 .04 a. Using Minitab, Excel, or other software that can determine probabilities for a binomial distribution with n=200 and p=.6, the exact answer is P(X140)=1P(X139)=1.9979=.0021. Since np =200(.6)=120 and np(1-p) =80 are both greater than 10 we can use the normal approximation. An approximation using the normal distribution is as follows: The mean and standard deviation are 140 120 2.89 so the np 200 (.6) 120 and np(1 p) 200 (.6)(1 .6) 6.928 . For 140, z 6.928 approximate answer is P(Z2.89) =1 P(Z<2.89)=1.9981=.0019. An as you can see this is pretty close to the exact probability (.0021) obtained from the binomial distribution. b. Note that 70% of 20 is 14 so the desired probability is P(X14). The exact answer using a binomial distribution with n= 20 and p=0.6 is P(X14)=1P(X13)=1.75=.25. In this case np =20(.6)=12 and np(1-p) =8 <10. So the conditions for using the normal approximation are not satisfied. However, since 8 is close to 10, if we proceed with the normal approximation we have the following: The mean and standard deviation are np 20(.6) 12 and np(1 p) 20 (.6)(1 .6) 2.191 . For 14, 14 12 0.91 so the approximate answer is P(Z0.91) =1 P(Z<0.91)=1.8186= .1814, which is not very close to 2.191 the exact probability (.25) that was obtained using the binomial distribution. z 9.2 The parameter of interest is the proportion that thinks crime is a serious problem in the population of all adult Americans. The statistic will be the proportion of the 1000 adults in the sample who think crime is a serious problem. The statistic will estimate the unknown value of the parameter. 9.17 a. Mean = p = .2 .2(1 .2) .05 . b. s.d. p̂ = 64 c. .15 and .25, calculated as .2 .05. d. .10 and .30, calculated as .2 (2.05). 9.18 .70 (1 .70 ) .0324 . 200 b. .70 (3.0324), or .6028 to .7972. c. pˆ 120 200 .60 . This is a statistic. d. The value .60 is slightly below the interval of possible sample proportions for 99.7% of all random samples of 200 from a population where p = .70. In other words, the sample proportion is unusually low if the true proportion were .70. Maybe the population value is actually less than .70. (This part is not graded.) a. Mean = .70; s.d. p̂ =