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1. Assume that 1 of the 100 subjects is randomly selected and find the
indicated probability.
Group A: 35 rh+ and 5rh- (blood)
Group B: 8rh+ and 2rhGroup AB: 4rh+ and 1rhGroup O: 39rh+ and 6rhthis question doesn’t ask for a probability – what is the indicated
probability of what??? – maybe #1 is just the set-up for number 2
2. If one of the subjects is randomly selected, find the probability of
getting someone who tests negative or someone who is not pregnant.
Positive test results (pregnancy is indicated) Negative test results
(pregnancy not indicated
prob(-) = (5+2+1+6)/100 = 0.14
prob(+) = (35+8+4+39)/100 = 0.86
so probability of getting + OR – is 1
What is this below????
Subject is pregnant 80 5
Subject not pregnant 3 11
3. Finding expected value in roulette. When you give a casino $5 for a
bet on the number 7 in roulette, you have a 1/38 probability of winning
$175 and a 37/38 probability of losing $5. If you bet $5 that the outcome
is an odd number, the probability of winning $5 is 18/38, and the
probability of losing $5 is 20/38.
A. If you bet $5 on number 7, what is you expected value?
Expected value = (1/38)(175) + (37/38)(-5) = -0.263
B. If you bet $5 that the outcome is an odd number, what is your expected
value?
Expected value = (18/38)(5) + (20/38)(-5) = = -0.263
C. Which of these options is best: bet on 7 bet on odd, or don’t bet?
don’t bet
4. Several students are unprepared for a multiple choice quiz with 10
questions, and all of their answers are guesses. Each question has five
possible answers, and only one of them is correct.
A. Find the mean and standard deviation for the number of correct answers
for such students.
Well, you could get 0 correct or 1 correct or 2…
So, (0+1+2+3+4+5+6+7+8+9+10) = 55, / 11 = 5 is the mean
s.d = sqrt ( (sum(1)^2 + (2)^2 + … + (10)^2 - (55)^2/11)/11) = 3.16
B. Would it be unusual for a student to pass by guessing and getting at least
7 correct answers? Why or why not?
Well, the chances of getting 77 correct would be (1/5)^7 = 0.0078, so the
chances are not that hight – so yes it would be unusual
5. Assume that the readings on the thermometers are normally distributed
with a mean of 0 degrees and s standard deviation of 1.00 degrees Celsius. A
thermometer is randomly selected and tested. Find the probability of Less
than -2.75.
look up in table and get 0.003
6. The lengths of pregnancies are normally distributed with a mean of 268
days and a standard deviation of 15 days.
A. One classic use of the normal distribution is inspired by a letter in which
a wife claimed to have given birth 308 days after a brief visit from her
husband, who was serving in the navy. Find the probability of a pregnancy
lasting 308 days or longer. What does the result suggest?
Z = (308 – 268)/15 = 2.67, look up in z-table to get 0.0038
So there is a 0.38% chance of this happening
She was probably cheeting on him!
B. If we stipulate that a baby is premature if the length of pregnant is in the
lowest 4%, find the length that separates premature babies from those who
are not premature. Premature babies often require special care, and this
result could be helpful to hospital administrators in planning for that care.
look up 0.04 for area under the normal curve in the table, to get a
z-value = -1.75 , X = (-1.75)(15) + 268 = 241.75 days
7. Sales of a telemarketer who worked four days are 1, 11, 9, 3. Assume that
samples of size 2 are randomly selected with replacement from this
population of four values.
A. List the 16 different possible samples and find the mean of each of them
this is assuming order is important:.
(1,1) (1,11) (1,9) (1,3) (11,11) (11,9) (11,3) (9,9) (9,3) (3,3)
(1,1) (11,1) (9,1) (3,1) (11,11) (9,11) (3,11) (9,9) (3,9) (3,3)
I disagree that there are 16 samples, (1,1) is not different from (1,1)…
B. Identify the probability of each sample, and then describe the sampling
distribution of sample means.
The probablility of ANY sample is 2/16, since (a,b) = (b,a)
The sampling distribution is a “box-shaped” distribution, since the
probability of them all is the same
C. Find the mean of the sampling distribution (from c) equal to the mean of
the population of the four listed values? Are those means always equal?
(1+11+9+3) / 4 = 6
no