College Stats Assignment #2 |

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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 *x*² *$ ** 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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