Download MATH 060, EXAM 3 (SOLUTIONS), 2 AUGUST, 2005 (1) A true

MATH 060, EXAM 3 (SOLUTIONS), 2 AUGUST, 2005
(1) A true-false test has 100 questions. Using the normal approximation to the binomial
distribution, estimate the probability that one will obtain a score of at least 60% by
answering the questions randomly.
The binomial distribution of interest
is B(100, 12 ), whose mean is (100)( 12 ) =
q
50 and whose standard deviation is (100)( 12 )( 12 ) = 5, and the corresponding
normal distribution we use to approximate is N(50, 5). We want P (X ≥ 60),
and this is equal to P (z ≥ 60−50
) = P (z ≥ 2) = .5 − P (0 ≤ z ≤ 2) =
5
(table 3 lookup) = .5 − .4772 = .0228.
(2) Consider the set {1, 2, 3} as a very small population, with each number having equal
likelihood of being picked in any random trial. List all the possible samples of size two
(i.e., one replaces the number after each trial), and draw a histogram for the sample
distribution of sample means.
There are 32 = 9 possible samples, (1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3);
only (1, 1) has mean x = 1, (1, 2) and (2, 1) have mean 1.5, (1, 3), (3, 1) and
(2, 2) have mean 2, (2, 3) and (3, 2) have mean 2.5, and (3, 3) has mean 3.
So mark off values 1, 1.5, 2, 2.5, and 3 on the x-axis; above 1 and 3 put a
bar of height 1/9; above 1.5 and 2.5 put a bar of height 2/9; and above 2
put a bar of height 3/9. There’s your histogram.
(3) The verbal portion of the SAT exam for female students was found to be normally distributed, with mean 500 and standard deviation 40. If you take a random sample of 100
students, what is the probability that the mean of their verbal SAT scores exceeds 512?
) = N(500, 4). P (X ≥ 512) =
The SDSM distribution is normal, N(500, √40
100
512−500
) = P (z ≥ 3) = (table 3 lookup) = .5 − .4987 = .0013.
P (z ≥
4
(4) A certain random variable is normally distributed, with an unknown mean µ and a
known standard deviation of 8. A random sample of size 16 is taken, and the sample
mean x is 200. What is the probability that 199 ≤ µ ≤ 201?
The SDSM is normal, N(µ, √816 ) = N(µ, 2). You’re looking for P (199 ≤ µ ≤
201) = P ( 199−200
≤ z ≤ 201−200
) = P (−.5 ≤ z ≤ .5) = (2)(P (0 ≤ z ≤ .5)) =
√8
√8
16
16
(table 3 lookup) = (2)(.1915) = .3830.
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(5) In Problem 4 above, suppose you want to double the probability that 199 ≤ µ ≤ 201.
What is the smallest sample size you can get away with to make this happen?
You’re looking for n large enough so that P (199 ≤ µ ≤ 201) = (2)(.3830) =
.7660. If the central portion of the bell curve has area .7660, then each tail
has area .1770. The corresponding z-value is z(.177) = 1.19 (obtainable from
√ , which
table 3); and the maximum error for sample size n is (1.19) √8n = 9.52
√n
we want to be ≤ 1. Thus n must be the smallest integer so that n ≥ 9.52;
i.e., n ≥ 90.63. Thus n = 91.
(6) The downtown post office advertises that the average waiting time at the service counter
is 4 minutes. You have reliable information that the standard deviation for the distribution of waiting times is 2 minutes, but you’re skeptical about the claimed mean.
You collect a sample of 25 post office customers, and find that the mean wait for these
people is 5 minutes. Find the 95% confidence interval for the mean waiting time. What
does this tell you about the advertised waiting time?
The sample size n = 25 justifies use of the central limit theorem, making
the SDSM, for all practical purposes, normal with unknown mean µ and
standard deviation √225 = .4. This makes our confidence interval equal to
x ± (.4)(z( 12 (1 − .95))) = 5 ± z(.025) = (table 3 lookup) = 5 ± .784 =
[4.216, 5.784]. The advertised mean of 4 lies below the lower limit of this
confidence interval, prompting one to conjecture that the post office claim is
bogus.
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