2 (x m1)2>2s1

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1 e s2 !2p

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2 (y m2)2>2s2 d

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and since this is a product of a function of x alone and a function of y alone for all values of x and y, it follows that X and Y are independent. Conversely, if X and Y are independent, f (x, y) given by (49) must for all values of x and y be the product of a function of x alone and a function of y alone. This is possible only if r 0.

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Miscellaneous distributions 4.52. Find the probability that in successive tosses of a fair die, a 3 will come up for the first time on the fifth toss.

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Method 1 The probability of not getting a 3 on the first toss is 5>6. Similarly, the probability of not getting a 3 on the second toss is 5>6, etc. Then the probability of not getting a 3 on the first 4 tosses is (5>6) (5>6) (5>6) (5>6) (5/6)4. Therefore, since the probability of getting a 3 on the fifth toss is 1>6, the required probability is 5 4 1 a b a b 6 6 625 7776

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Method 2 (using formula) Using the geometric distribution, page 117, with p bility is

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1 5 4 a ba b 6 6

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1>6, q

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625 7776

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5>6, x

5, we see that the required proba-

4.53. Verify the expressions given for (a) the mean, (b) the variance, of the Weibull distribution, page 118.

E(X)

b 30 abx e u

ab ` u a be a1>b 30 a

1 (1>b) u b du

1>b 30 u e

1 b b

where we have used the substitution u

axb to evaluate the integral.

30 abx

b 1 e axb dx (1>b)

E(X2)

ab ` u 1 a b a1>b 30 a

1 (1>b) u b

1 du

30 u

2>b e u du

2 b b

Then

s2 E[(X a

2>b c

m)2] a1

E(X2) 2 b b

2 a1

1 bd b

CHAPTER 4 Special Probability Distributions

Miscellaneous problems 4.54. The probability that an entering college student will graduate is 0.4. Determine the probability that out of 5 students (a) none, (b) 1, (c) at least 1, will graduate.

(a) P(none will graduate) (b) P(l will graduate)

0 5 5C0(0.4) (0.6)

0.07776, or about 0.08

1 4 5C1(0.4) (0.6)

0.2592, or about 0.26 0.92224, or about 0.92

(c) P(at least 1 will graduate)

P(none will graduate)

4.55. What is the probability of getting a total of 9 (a) twice, (b) at least twice in 6 tosses of a pair of dice

Each of the 6 ways in which the first die can fall can be associated with each of the 6 ways in which the second die can fall, so there are 6 6 36 ways in which both dice can fall. These are: 1 on the first die and 1 on the second die, 1 on the first die and 2 on the second die, etc., denoted by (1, 1), (1, 2), etc. Of these 36 ways, all equally likely if the dice are fair, a total of 9 occurs in 4 cases: (3, 6), (4, 5), (5, 4), (6, 3). Then the probability of a total of 9 in a single toss of a pair of dice is p 4 >36 1 >9, and the probability of not getting a total of 9 in a single toss of a pair of dice is q 1 p 8 >9. (a) P(two 9s in 6 tosses) (b) P(at least two 9s)

6C2 a

1 2 8 6 b a b 9 9

61,440 531,441

P(four 9s)

P(two 9s)

6C2 a 9 b

P(three 9s)

4 6C3 a 9 b

P(five 9s) 1

P(six 9s)

6C5 a 9 b

8 a b 9

8 a b 9

6C4 a 9 b

8 a b 9

6C6 a 9 b

61,440 531,441 Another method P(at least two 9s)

10,240 531,441

960 531,441

48 531,441

1 531,441

72,689 531,441

P(zero 9s)

6C0 a 9 b

P(one 9)

6C1 a 9 b

8 6 a b 9

8 5 a b 9

72,689 531,441

4.56. If the probability of a defective bolt is 0.1, find (a) the mean, (b) the standard deviation for the distribution of defective bolts in a total of 400.

(a) Mean (b) Variance np 400(0.1) npq 40, i.e., we can expect 40 bolts to be defective. 36. Hence the standard deviation !36 6.

400(0.l)(0.9)

4.57. Find the coefficients of (a) skewness, (b) kurtosis of the distribution in Problem 4.56.

(a) Coefficient of skewness q p 0.9 6 1 0.1 ! npq 6pq npq 0.133

Since this is positive, the distribution is skewed to the right. (b) Coefficient of kurtosis 3 1 3 6(0.1)(0.9) 36 3.01

The distribution is slightly more peaked than the normal distribution.

4.58. The grades on a short quiz in biology were 0, 1, 2, . . . , 10 points, depending on the number answered correctly out of 10 questions. The mean grade was 6.7, and the standard deviation was 1.2. Assuming the grades to be normally distributed, determine (a) the percentage of students scoring 6 points, (b) the maximum grade of the lowest 10% of the class, (c) the minimum grade of the highest 10% of the class.

(a) To apply the normal distribution to discrete data, it is necessary to treat the data as if they were continuous. Thus a score of 6 points is considered as 5.5 to 6.5 points. See Fig. 4-20. 5.5 in standard units 6.5 in standard units (5.5 (6.5 6.7)>1.2 6.7)>1.2 1.0 0.17