CIS321 Fall, 1999

30. Suppose that the probabilities are 0.4, 0.3, 0.2, and 0.1 that there will be 0, 1, 2, or 3 power failures in a certain city during the month of July. Use the formulas which define m and s ² to find

1. the mean of this probability distribution;
2. the variance of this probability distribution.

(a) m = 0´ 0.4 + 1´ 0.3 + 2´ 0.2 + 3´ 0.1 = 1

(b) s ² = (0-1)^2´ 0.4 + (1-1)^2´ 0.3 + (2-1)^2´ 0.2 + (3-1)^2´ 0.1 = 1

4.34 Find the mean and the variance of the uniform probability distribution given by

f(x) = 1/n for x=1,2,3,…,n

m = å xf(x) = å x´ (1/ n) =(1/ n) å x = (1/n)´ [n(n+1)/2] = (n+1)/2

s ² = å (x-m )² f(x) = å (x-m )² ´ (1/ n) = (1/ n) å (x-m )²

= (1/ n) (å x² + 2m å x + å m ²)

= (1/ n){[1/6´ n(n+1)(2n+1)]-[2 ´ (n+1)/2 ´ n(n+1)/2] + n ´ [(n+1)/2]²}

= [(n+1)(2n+1)]/6 – (n+1)²

= (n²-1)/12

4.38 If 95% of certain high performance radial tires last at least 30,000 miles, find the mean and the standard deviation of the distribution of the number of these tires, among 20 selected at random, that last at least 30,000 miles, using

1. Table 1, the formula which defines m , and the computing formula for s ²;

(b) the special formulas for the mean and the variance of a binomial distribution.

1. b(x; n, p) = B(x; n, p) – B(x-1; n, p)

By look up Table1, we get

b(1; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(2; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(3; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(4; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(5; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(6; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(7; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(8; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b( 9; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(10; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(11; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(12; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(13; 20, 0.95) = 0.0000 – 0.0000 = 0.0000

b(14; 20, 0.95) = 0.0003 – 0.0000 = 0.0003

b(15; 20, 0.95) = 0.0026 – 0.0003 = 0.0023

b(16; 20, 0.95) = 0.0159 – 0.0026 = 0.0133

b(17; 20, 0.95) = 0.0755 – 0.0159 = 0.0596

b(18; 20, 0.95) = 0.2642 – 0.0755 = 0.1887

b(19; 20, 0.95) = 0.6415 – 0.2642 = 0.3773

b(20; 20, 0.95) = ₂₀C₂₀ (0.95)²⁰ (1-0.95)⁰ = 0.3585

m = å xf(x) = å xb(x; 20, 0.95) for x=0,1,2,3,… 20.

= 14´ 0.0003+15´ 0.0023+16´ 0.0133+17´ 0.0596+18´ 0.1887+19´ 0.3773+20´ 0.3585

= 19

s ² = å (x-m )² f(x) = å (x-19)² ´ b(x; 20, 0.95)

=(14-19)^2´ 0.0003+(15-19)² ´ 0.0023+(16-19)^2´ 0.0133+(17-19)^2´ 0.0596

+(18-19)^2´ 0.1887+(19-19)^2´ 0.3773+(20-9)^2´ 0.3585

=0.0075+0.0368+0.1197+0.2384+0.1884+0.3585

= 0.9493

2. m = n ´ p = 20 ´ 0.95 = 19

s ² = n ´ p ´ (1-p) = 20 ´ 0.95 ´ (1-0.95) = 0.95

4.42 Construct a table showing the upper limits provided by Chebyshev’s theorem for the probabilities of obtaining values differing from the mean by at least 1, 2, and 3 standard deviations and also the corresponding probabilities for the binomial distribution with n=16 and p=1/2.

P( |X-m | ³ ks ) £ 1 / k²

m = n ´ p = 16 ´ ½ = 8

s ² = n ´ p ´ (1-p) =16 ´ 1/2 ´ (1-1/2) = 4

s = 2

Substitute m =8 , s = 2 into the formula |X-m | ³ ks , we get

when k=1, P(x £ 6 or x ³ 10) £ 1

when k=2, P(x £ 4 or x ³ 12) £ 1/4

when k=3, P(x £ 2 or x ³ 14) £ 1/9

Using Table 1, we get

when k=1, x £ 6, B(6; 16, ½) = 0.2272

when k=2, x £ 4 , B(4; 16, ½) = 0.0384

when k=3, x £ 2 , B(2; 16, ½) = 0.0021

As p=1/2, the binomial distribution is symmetric.

when k=1, x £ 6 or x ³ 10, the Upper limits of probability is 2´ 0.2272 = 0.4544

when k=2, x £ 4 or x ³ 12, the Upper limits of probability is 2´ 0.0384 = 0.0768

when k=3, x £ 2 or x ³ 14, the Upper limits of probability is 2´ 0.0021 = 0.0042

From the above table, we notice that the upper limits given by Chebyshev’s Theorem is very large compared to the true probability values calculated out from binomial distribution. The results above are consistent with the statement from the textbook, namely "Chebyshev’s theorem applies to any probability distribution for which m and s exist. However, it provides only an upper limit (often a very poor one) to the probability." In order to get a more precise result, we need to take advantage of the features of specific distribution given (e.g. binomial distribution in this question).

4.46 How many times do we have to flip a balanced coin to be able to assert with a probability of at most 0.01 that the difference between the proportion of tails and 0.50 will be at least 0.04?

P( |X-m | ³ ks ) £ 1 / k²

m = n ´ p

s ² = n ´ p ´ (1-p)

From the statement of the question, we get

1. 1/ k²= 0.01

(b) "the difference of number of tails and 0.5n is at least 0.04n."

So, (c) k=10

(d) m = n ´ p =0.5n, therefore, p = n / 2

(e) ks = 10s =0.04n,

(f) s ² = n ´ p ´ (1-p) = m ´ (1-p) = n / 2 ´ (1-1/2) = n / 4

Solve the equation of s ² = n / 4 and 10s =0.04n , we get n =100/0.08² = 15625.

53. In a given city, 6% of all drivers get at least one parking ticket per year. Use the Poisson approximation to the binomial distribution to determine the probabilities that among 80 drivers (randomly chosen in this city)

1. four will get at least one parking ticket in any given year;
2. at least three will get at least one parking ticket in any given year;
3. anywhere from three to six, inclusive, will get at least one parking ticket in any given year.

1. l = np = 80´ 6% = 4.8

f(4; 4.8) = F(4; 4.8) – F(3; 4.8) = 0.476-0.294 = 0.182

2. 1 – F(2; 4.8) = 1-0.143 = 0.857

(c) F(6; 4.8) – F(2; 4.8) = 0.791 – 0.143 = 0.648

4.57 At a checkout counter customers arrive at an average rate of 1.5 per minute. Find the probabilities that

1. at most four will arrive in any given minute;

(b) at least three will arrive during an interval of 2 minutes;

(c) at most 15 will arrive during an interval of 6 minutes.

1. F(4; 1.5) = 0.981

(b) l = 1.5 ´ 2 = 3

1-F(2; 3) = 1 – 0.423 = 0.577

(c) l = 1.5 ´ 6 = 9

F(15; 9) = 0.978

4.60 An expert shot hits a target 95% of the time. What is the probability that the expert will miss that target for the first time on the fifteenth shot?

p = 1- 95% = 0.05

x = 15

g(15; 0.05) = (0.05) ´ (1-0.05))^15-1 = 0.024

The probability that the expert will miss that target for the first time on the fifteenth shot is 0.024.

4.74 Suppose that the probabilities are, respectively, 0.60, 0.20, 0.10, and 0.10 that an income tax form will be filled in correctly, that it will contain an error favoring the taxpayers, that it will contain an error favoring the government, or that it will contain both kinds of errors. Find the probability that among 10 or the income tax forms randomly selected for audit five will be correct, three will contain an error favoring the taxpayer, one will contain an error favoring the government, and one will contain both kinds of errors.

Substituting n=10, x1=5, x2=3, x3=1 x4=1, p1=0.60, p2=0.20, p3=0.10 and p4=0.10 into the formula of multinomial distribution, we get

f(5, 3, 1, 1) = [10!/(5! 3! 1! 1!)] (0.60)⁵ (0.20)³ (0.10)¹ (0.10)¹ = 0.0312

4.70 Poisson probabilities can be calculated using the MINITAB command

PDF;

POISSON mean = 1.64.

Which produces the output

POISSON WITH MEAN= 1.640

K P( X = K)

0 0.1940

1 0.3181

2 0.2609

3 0.1426

4 0.0585

5 0.0192

6 0.0052

7 0.0012

8 0.0003

9 0.0000

Find the Possion probabilities with l = 2.73.

Worksheet size: 100000 cells

Probability Density Function

Poisson with mu = 2.73000

x P( X = x)

0 0.0652

1 0.1780

2 0.2430

3 0.2212

4 0.1509

5 0.0824

6 0.0375

7 0.0146

8 0.0050

9 0.0015

10 0.0004

11 0.0001

12 0.0000