1. If the population is normally distributed then the sample mean is also normally distributed even for small sample size 2. The level of significance indicates the probability of rejecting a false null hypothesis. 3. An estimator is called consistent if its variance and standard deviations consistently remain the same regardless of changes in the sample size 4. We do not need to perform the continuity correction if the population is 20 times or more than the sample size 5. When the level of confidence and sample standard deviation remain the same, a confidence interval for a population mean based on a sample of n=100 will be narrower than a confidence interval for a population mean based on a sample of n=50. 6. For a continuous distribution, the exact probability of a particular value is always zero. 7. A fastener manufacturing company uses a chi-square goodness of fit test to determine if a population of all lengths of ¼ inch bolts it manufactures is distributed according to a normal distribution. If we reject the null hypothesis, it is reasonable to assume that the population distribution is at least approximately normally distributed. Solution 1. True 2. True 3. True 4. False 5. False 6. False 7. True.

1. 1. If the random variable z is the standard normal score, is it true that P(z < 3) could easily be
approximated without referring to a table? Why or why not? yes , because if you plug i 1,0,2 it
can be true 2. Given a binomial distribution with n = 35 and p = 0.86, would the normal
distribution provide a reasonable approximation? Why or why not? it would not provide a
reasonable approximation. this is because n<26(1-0.86)=3.64<5 3. Find the area under the
standard normal curve for the following: (A) P(z > 1.15) (B) P(0 < z < 1.58) (C) P(-1.06 < z <
1.25) a. p(z > 1.15) = 0.12507 b.p(0 < z < 1.58) = .94295 - .5 = .44295 c . p(-1.06 < z 1.15)=
.87493 - .14457 = .73036 4. Find the value of z such that approximately 9.48% of the
distribution lies between it and the mean. 9.48 turns into .0948 0.5000+ 0.0948=0.5948 0.5000-
0.0948=0.4042 z=0.2399 z=-0.2399 5. Assume that the average annual salary for a worker in the
United States is $33,000 and that the annual salaries for Americans are normally distributed with
a standard deviation equal to $6,250. Find the following: (A) What percentage of Americans earn
below $21,000? (B) What percentage of Americans earn above $41,000? a. 21,000-33000=8000
8000/62000=0.193548387 p(x<21,000)=p(z<-0.193548)=0.0287 b.41000-33000=8000
8000/62000=0.129032258 p(x<41,000)=p(z<0.129032258)=0.5478 6. X has a normal
distribution with a mean of 80.0 and a standard deviation of 3.5. Find the following probabilities:
(A) P(x < 74.0) (B) P(76.0 < x < 85.0) (C) P(x > 89.0) A.z=74.0-80=-6 z=-6/3.5=-1.714285714
z=-1.714285714 p(z<74.0)=p(x<1.714285714)=0.9564 b.76-80=-4 -4/3.5=-1.142857143 85-
80=5 5/3.5=1.428571429 p(7689)=p(z>2.571428572)=.09949 7. Answer the following: (A) Find
the binomial probability P(x = 5), where n = 12 and p = 0.30. (B) Set up, without solving, the
binomial probability P(x is at most 5) using probability notation. (C) How would you find the
normal approximation to the binomial probability P(x = 5) in part A? Please show how you
would calculate
Solution
1)