College Stats Assignment #2

1/ Convert each of the following probabilities to odds:

(a) The probability of rolling 10 or less with a pair of balanced dice is 11/12.
(b) If a researcher randomly selected 5 of 30 homes to be included in a study, the probability is 1/6 that any particular home will be included.
(c) The probability of getting at least 2 heads in five flips of a balanced coin is 13/16.

(6)

2/ Convert each of the following odds to probabilities:

(a) The odds are 19 to 5 that a given horse will not win the Kentucky Derby.
(b) If 5 cards are drawn with replacement from an ordinary deck of 52 playing cards,
the odds are 13 to 3 that at most 3 of the cards will be red.
(c) If 2 persons are chosen at random from a group of 10 men and 12 women,
the odds are 40 to 37 that one man and one woman will be selected.

(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.

(a) Find the mean of this probability distribution.
(b) use the computing formula to find its variance.

(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:

(a) exactly 14 have a color TV;
(b) at least 14 have a color TV;
(c) at most 14 have a color TV;

(5)

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

(a) neither coin is counterfeit;
(b) one of the coins is counterfeit;
(c) both coins selected are counterfeits?

(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:

(a) between z = 0 and z = 0.87;
(b) between z = – 1.66 and z = 0;
(c) to the right of z = 0.48;
(d) to the right of z = – 0.27;
(e) to the left of z = 1.30;
(f) to the left of z = – 0.79;
(g) between z = 0.55 and z = 1.12;
(h) between z = – 1.05 and z = – 1.75;
(i) between z = – 1.95 and z = 0.44;
(j) to the left of z = – 0.15.

(10)

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

(a) between 0 and z is 0.4726;
(b) to the left of z is 0.9868;
(c) to the left of z is 0.3085;
(d) between – z and z is 0.8502;
(e) between 0 and z is 0.4306;
(f) to the right of z is 0.7704;
(g) to the right of z is 0.1314;
(h) between – z and z is 0.9700.

(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)

.

.