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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(X140)=1P(X139)=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(Z2.89) =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(X14). The exact answer using a binomial distribution
with n= 20 and p=0.6 is P(X14)=1P(X13)=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(Z0.91) =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  =