College Stats Assignment #2

Show your work!

1/ Convert each of the following probabilities to odds:

(6)

2/ Convert each of the following odds to probabilities:

(6)

3/ If a contractor feels that 17 to 8 are fair odds that a job will be finished on time,
what probability does he assign to the event?

(3)

4/ In each case, determine whether the given values can be looked upon as the values
of a probability distribution of a random variable x if the only
values for x are 1, 2, 3, 4, and 5. Explain your answers completely

(a) f(1) = 0.18 f(2) = 0.20 f(3) = 0.22 f(4) = 0.20 f(5) = 0.18;

(b) f(1) = 0.05 f(2) = 0.05 f(3) = 0.05 f(4) = – 0.05 f(5) = 0.90;

(c) f(1) = 0.07 f(2) = 0.23 f(3) = 0.50 f(4) = 0.19 f(5) = 0.01;

(6)

5/ The probabilities are 0.22, 0.34, 0.25, 0.13, 0.05, and 0.01 that 0, 1, 2, 3, 4, or 5 of a doctor's patients will come down with the flu during the first week of January.

(4)

6/ Using the formulae, find µ and s for a binomial distribution with n = 9 and p = 0.50.

(2)

7/ If 95% of the families surveyed in Sept-Iles have a color TV set,
use the Binomial Table to find the probabilities that among 16 such families:

(5)

8/ If 4 of 18 antique gold coins are counterfeits and 2 of them are randomly selected,
what are the probabilities that:

(6)

9/ According to Chebyshev's theorem, with what probability can we assert that
in 40,000 flips of a balanced coin the proportion of heads will be
between 0.45 and 0.55? (hint: find µ and
s for the distribution with n = 40,000 and p = ?)

(4)

10/ Find the standard-normal curve area which lies:

(10)

11/ Find z if the standard-normal-curve area

(8)

TOTAL (60)

Solutions

1/

a) odds are 11 to 1 of getting 10 or less.

b) odds are 5 to 1 that the home is not included.

c) odds are 13 to 3 of getting at least 2 heads.

(6)

2/

a) Since P(win) = 5/24, then P(not win) = 1 – P(W) = 19/24 = 0.7917

b) P(Red) = 13/16 = 0.813

c) P(1 man, 1 woman) = 40/77 = 0.52

(6)

3/ P(F) = 17/25 = 0.68

(3)

4/

a) no, sum of probabilities = 0.98 ! 1

b) no, f(4) < 0

c) yes, both conditions are satisfied.

(6)

5/

a) l = S x$ f(x) = 1.48

b) r ² = [S x² $ f(x)] – l² = 1.37

(4)

6/

(2)

7/

a) P(x = 14) = 0.146

b) P(x m 14) = P(14) + P(15) + P(16) = 0.146 + 0.371 + 0.44 = 0.957

c) P(x [ 14) = 1 – [P(15) + P(16)] = 1 – [0.371 + 0.44] = 0.189

(5)

8/ The total number of ways to pick 2 of 18 coins is 18 C 2 = 153

a) We must choose our 2 non-counterfeits from the 14 good coins
so there are
14 C 2 = 91 ways to do that.
The probability of choosing 2 good coins = 91/153 =
0.595

b) Now we need the number of ways to choose 1 of 4 counterfeits and 1 of 14 good coins.
This is
4 C 1 $ 14 C 1 = 56 , so the probability of choosing 1 good and 1 bad is 56/153 = 0.366

c) The number of ways to choose 2 of 4 counterfeits is 4 C 2 = 6,
so probability =
6/153 = 0.0392

(6)

9/ The probability of Heads is 0.5. This is a proportion situation (Binomial)
so
l = np = 20,000; and r = [np(1 – p)] 1/ 2 = 100
k is the number of standard deviations between X and
l
if p lies between 0.45 and 0.55, then k = [ 0.55(40,000) – 20,000 ] divided by r = 20
The theorem says the probability is
at least (1 – 1/k²) = 1 – 1/400 = 0.9975

(4)

10/

a) b)

c) d)

e) f)

g) h)

i) j)

(10)

11/

a) b)

c) d)

e) f)

g) h)

(8)

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